Equations of Tangent and Normal to a Parabola || Exercise

Exercise

1. Find the equation of tangent and normal to the parabola
   1. \( y^2=8x\) at (2,-4)
      

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( y^2 = 4ax \), so we have \( 4a = 8 \), so \( a = 2 \)
         Also, the slope of tangent to the parabola at \( (x_1, y_1)=(2,-4) \) is
         \( m_1 = \frac{2a}{y_1} =\frac{2.2}{-4}=-1\)
         Now, equation of tangent to the parabola is
         \( y-y_1=m_1(x-x_1)\)
         \( y+4 = (-1) (x - 2) \)
         or\( y+4=-x+2 \)
         or\( x + y + 2 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, slope of normal to the parabola at \( (x_1, y_1)=(2,-4) \) is
         \( m_2 = \frac{y_1}{2a} =1\)
         Now, equation of normal to the parabola is
         \( y-y_1=m_2(x-x_1)\)
         \( y+4 = (1) (x - 2) \)
         or\( y+4=x-2 \)
         or\( x - y -6= 0 \) Answer: normal line
   2. \( x^2=12y\) at (-6,3)

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( x^2 = 4ay \), so we have \( 4a = 12 \), so \( a = 3 \)
         Also, the slope of tangent to the parabola at \( (x_1, y_1) = (-6, 3) \) is
         \( m_1 = \frac{x_1}{2a} =\frac{-6}{2 \cdot 3} = -1 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 3 = (-1) (x + 6) \)
         or\( y - 3 = -x - 6 \)
         or\( x + y + 3 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, the slope of the normal to the parabola at \( (x_1, y_1) = (-6, 3) \) is
         \( m_2 = -\frac{1}{m_1} = 1 \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 3 = (1) (x + 6) \)
         or\( y - 3 = x + 6 \)
         or\( x - y + 9 = 0 \) Answer: normal line
   3. \( x^2=4y\) at the point whose abscissa is 6

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( x^2 = 4y \), so we have \( 4a = 4 \), so \( a = 1 \)
         Given \( x_1 = 6 \), we find the corresponding \( y_1 \) by substituting into the equation:
         \( y_1 = \frac{x_1^2}{4} = \frac{6^2}{4} = 9 \)
         So, the point of tangency is \( (6, 9) \).
         Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (6, 9) \) is
         \( m_1 = \frac{x_1}{2a} = \frac{6}{2 \cdot 1} = \frac{6}{2} =3 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 9 = 3(x - 6) \)
         or\( y - 9 = 3x - 18 \)
         or\( 3x - y -9 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, the slope of the normal to the parabola at \( (x_1, y_1) = (6, 9) \) is
         \( m_2 = -\frac{1}{m_1} = -\frac{1}{3} \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 9 = -\frac{1}{3}(x - 6) \)
         or\( 3y - 27 = -x + 6 \)
         or\( x +3 y - 33 = 0 \) Answer: normal line
   4. \( y^2=16ax\) at the point whose ordinate is -4a

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( y^2 = 16ax \).
         Given \( y_1 = -4a \), we find the corresponding \( x_1 \) by substituting into the equation:
         \( (-4a)^2 = 16a \cdot x_1 \implies 16a^2 = 16a \cdot x_1 \implies x_1 = a \)
         So, the point of tangency is \( (a, -4a) \).
         Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (a, -4a) \) is
         \( m_1 = \frac{8a}{y_1} = \frac{8a}{-4a} = -2 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y + 4a = -2(x - a) \)
         or\( y + 4a = -2x + 2a \)
         or\( 2x + y + 2a = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         The slope of the normal to the parabola at \( (x_1, y_1) = (a, -4a) \) is
         \( m_2 = \frac{-1}{m_1} = \frac{1}{2} \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y + 4a = \frac{1}{2}(x - a) \)
         or\( 2(y + 4a) = x - a \)
         or\( x - 2y - 9a = 0 \) Answer: normal line
   5. \( y^2=12x\) at each end of the latus rectum

      Answer 👉 Click Here

      We know that, end of the latus rectum are \(a,\pm 2a)\).
      For the parabola \( y^2 = 12x \), we have \( 4a = 12 \), so \( a = 3 \). The end of latus rectum are \( (3, 6) \) and \( (3, -6) \).
      

      1. At the End \( (3, 6) \)
         Equation of the Tangent
         The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, 6) \) is
         \( m_1 = \frac{y_1}{2a} = \frac{6}{6} = 1 \)
         Now, the equation of the tangent is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 6 = 1(x - 3) \)
         or\( y - 6 = x-3 \)
         or\( x-y+3=0 \) Answer: tangent line
         Equation of the Normal
         The slope of the normal is the negative reciprocal of the tangent’s slope:
         \( m_2 = -\frac{1}{m_1} = -1 \)
         Now, the equation of the normal is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 6 = -1(x - 3) \)
         or\( y - 6 = -x + 3 \)
         or\( x + y - 9 = 0 \) Answer: normal line
      2. At the End \( (3, -6) \)
         
         
         The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, -6) \) is
         \( m_1 = \frac{y_1}{2a} = \frac{6}{-6} = -1 \)
         Now, the equation of the tangent is
         \( y - y_1 = m_1(x - x_1) \)
         \( y + 6 = -1(x - 3) \)
         or\( y + 6 = -x+3 \)
         or\( x+y+3=0 \) Answer: tangent line
         Equation of the Normal
         The slope of the normal is the negative reciprocal of the tangent’s slope:
         \( m_2 = -\frac{1}{m_1} = 1 \)
         Now, the equation of the normal is
         \( y - y_1 = m_2(x - x_1) \)
         \( y + 6 = 1(x - 3) \)
         or\( y + 6 = x - 3 \)
         or\( x-y -9 = 0 \) Answer: normal line
2. Solve the followings
   1. Prove that the line \(3x+4y+6=0\) is tangent to the parabola \(2y^2=9x\) and find its point of contact
      

      Answer 👉 Click Here

      
      The given parabola is \(2y^2 = 9x\), which is
      \( y^2 = \frac{9}{2} x \) (i)
      Thus,
      \( a = \frac{9}{8} \)
      The given line is
      \(3x + 4y + 6 = 0\)
      or \( y = -\frac{3}{4}x - \frac{3}{2} \) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(- \frac{3}{2} =\frac{\frac{9}{8}}{ -\frac{3}{4}}\) TRUE
      So, the line (ii) is tangent to the parabola (i)
      Next, to find the point of contact, we substitute \(x = -\frac{4y + 6}{3}\) into the parabola , so that
      \( 2y^2 = 9 \left(-\frac{4y + 6}{3}\right) \)
      \( 2y^2 = -12y - 18 \)
      \( 2y^2 + 12y + 18 = 0 \)
      \( y = -3 \)
      Substitute \( y = -3 \) into the line equation \(3x + 4y + 6 = 0\) , we get
      \( 3x + 4(-3) + 6 = 0 \)
      \( x=2 \)
      So, the point of contact is \( (2, -3) \)
   2. Prove that the line \(lx+my+n=0\) touches the parabola \(y^2=4ax\) if \(ln =am^2\)
      

      Answer 👉 Click Here

      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(lx+my+n=0\)
      or \( y = -\frac{l}{m}x - \frac{n}{m} \) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(- \frac{n}{m} =\frac{a}{ -\frac{l}{m}}\) TRUE
      \(ln =am^2\)
   3. For what value of a, will the straight line \(y=2x+3\) touch the parabola \(y^2=4ax\)
      

      Answer 👉 Click Here

      
      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(y=2x+3\) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(3 =\frac{a}{2}\) TRUE
      \(a=6\)
   4. If the line \(2x+4y=3\) touches the parabola \(y^2=4ax\) find the length of the latus rectum
      

      Answer 👉 Click Here

      
      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(2x+4y=3\)
      \(y=-\frac{1}{2}x+\frac{3}{4}\) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(\frac{3}{4} =\frac{a}{-\frac{1}{2}}\) TRUE
      \(a=-\frac{3}{8}\)
      Now, the length of rectum is
      \(4a=4 \times \frac{3}{8}=\frac{3}{2} \)
3. Solve the followings
   1. Find the equation of tangent to the parabola \(y^2=4x\)
      (i) parallel to the line \(x-2y+6=0\).
      (ii) perpendicular to the line \(2x-y=4\)
      Also find the point of contact.

      Answer 👉 Click Here

      (i) parallel to the line \(x-2y+6=0\). Also find the point of contact.
      

      
      The given parabola is
      \(y^2=4x\) (i)
      Thus,
      \( 4a = 4\) gives \(a=1\)
      The equation of tangent to parabola, which is parallel to \( x - 2y + 6 = 0 \) is
      \( x - 2y + k = 0 \) (A)
      \( 2y = x + k \)
      \( y = \frac{1}{2}x +\frac{k}{2} \) (ii)
      The condition for the line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(\frac{k}{2} =\frac{1}{\frac{1}{2}}\)
      \(k=4\)
      Hence, the required equation of tangent line is, from (A)
      \( x - 2y + 4 = 0 \)
      Next, to find the point of contact, we substitute \(x = 2y - 4\) into the parabola , so that
      \( y^2 = 4 \left(2y - 4 \right) \)
      \( y^2 = 8y - 16 \)
      \( y^2 - 8y +16 = 0 \)
      \( y = 4 \)
      Substitute \( y = 4 \) into the line equation \(x - 2y + 4 = 0\) , we get
      \( x - 2(4) + 4 = 0\)
      \( x=4 \)
      So, the point of contact is \( (4,4) \)

      ---

      (ii) perpendicular to the line \(2x-y=4\)
      Also find the point of contact.

      
      The given parabola is
      \(y^2=4x\) (i)
      Thus,
      \( 4a = 4\) gives \(a=1\)
      The equation of tangent to parabola, which is perpendicular to \(2x-y=4\) is
      \( x + 2y + k = 0 \) (A)
      \( 2y = -x - k \)
      \( y = -\frac{1}{2}x -\frac{k}{2} \) (ii)
      The condition for the line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(-\frac{k}{2} =\frac{1}{-\frac{1}{2}}\)
      \(k=4\)
      Hence, the required equation of tangent line is, from (A)
      \( x + 2y + 4 = 0 \)
      Next, to find the point of contact, we substitute \(x = - 2y - 4\) into the parabola , so that
      \( y^2 = 4 \left(-2y - 4 \right) \)
      \( y^2 = -8y - 16 \)
      \( y^2 + 8y +16 = 0 \)
      \( y =- 4 \)
      Substitute \( y = -4 \) into the line equation \(x + 2y + 4 = 0\) , we get
      \( x + 2(-4) + 4 = 0\)
      \( x=4 \)
      So, the point of contact is \( (4,-4) \)
   2. Find the rquation of normal line to the parabola \(y^2=3x\)
      (i) parallel to the line \(y=2x+1\)
      (ii) perpendicular to the line \(3x+2y=8\)
      

      Answer 👉 Click Here

      (i) parallel to the line \(y=2x+1\)
      

      
      The given parabola is
      \(y^2=3x\) (i)
      Thus,
      \( 4a = 3\) gives \(a=\frac{3}{4}\)
      The equation of normal to parabola, which is parallel to \(y=2x+1\) is
      \( y=2x+k \) (A)
      The condition for the line (A) to be normal to the parabola (i) is
      \(c=-2am-am^3\)
      \(k=-2\frac{3}{4}. 2-\frac{3}{4} (8)\)
      \(k=-9\)
      Hence, the required equation of tangent line is, from (A)
      \( y=2x-9 \)
      

      ---

      (ii) perpendicular to the line \(3x+2y=8\)
      

      
      The given parabola is
      \(y^2=3x\) (i)
      Thus,
      \( 4a = 3\) gives \(a=\frac{3}{4}\)
      The equation of normal to parabola, which is perpendicular to \(3x+2y=8\) is
      \( 2x-3y+k=0 \) (A) \( 3y=2x+k \)
      \( y=\frac{2}{3}x+\frac{k}{3} \) (ii)
      The condition for the line (ii) to be normal to the parabola (i) is
      \(c=-2am-am^3\)
      \(\frac{k}{3} =-2\frac{3}{4}. \frac{2}{3} -\frac{3}{4} (\frac{8}{27})\)
      \(k=-\frac{11}{3}\)
      Hence, the required equation of tangent line is, from (A)
      \( 3y=2x-\frac{11}{3} \)
      \( 9y=6x-11\)
      
4. Solve the followings
   1. Show that tangent to the parabola \(y^2=4x\) and \(x^2=4y\) at (1,2) and (-2,1) respectively are at right angles.
      
      
      Slope of the tangent to the Parabola \( y^2 = 4x \) at \( (1, 2) \)is
      \( m_1 = \frac{y_1}{2a} \)
      \( m_1 = \frac{2}{2} = 1 \)
      Next, slope of the tangent to the Parabola \( x^2 = 4y \) at \( (-2, 1) \)is
      \( m_2 = \frac{x_1}{2a} \)
      \( m_2 = \frac{-2}{2} = -1 \)
      The tangents are perpendicular if the product of their slopes is \(-1\).
      \( m_1 \times m_2 = 1 \times -1 = -1 \)
      Since the product of the slopes is \(-1\), the tangents are at right angles
   2. Show that normal to the parabola \(y^2=8x\) at (2,4) meets the parabola again in (18,-12)
      
      
      The equation of the parabola is \( y^2 = 8x \), so the slope of normal to the parabola is
      \( m= -\frac{2a}{y_1} \)
      \( m= - \frac{2.2}{4} = -1 \)
      Now, equation of the Normal at \( (2, 4) \) is
      \( y - y_1 = m(x - x_1) \)
      \( y - 4 = -1(x - 2) \)
      \( y = -x + 6 \)
      \( x + y - 6 = 0 \)
      To find where the normal meets the parabola again, substitute the equation of the normal \( x + y - 6 = 0 \) into the equation of the parabola \( y^2 = 8x \), we get
      \( y^2 = 8x \)
      \( (6 - x)^2 = 8x \)
      \( 36 - 12x + x^2 = 8x \)
      \( x^2 - 20x + 36 = 0 \)
      \( x = 2 \) or \( x = 18 \)
      When \( x = 2 y = 4 \) is the original point, and when \( x = 18 \), we get \( y = 6 - 18 = -12 \).
      Therefore, the normal meets the parabola again at \( (18, -12) \)
5. Solve the followings
   1. Find the equation of tangent to the parabola \(y^2=6x\) making angle 45 degree with the x-axis. Also find the point of contact.
      
      
      Given that, parabola is \(y^2=6x\), thus,
      \( 4a = 6\) gives \(a=\frac{3}{2}\)
      Also, given that tangent makes an angle of 45 degrees with the x-axis. Therefore, the slope \(m\) of the tangent is
      \( m = \tan(45^\circ) = 1 \)
      Now, the equation of the tangent to the parabola \( y^2 = 4ax \) is
      \( y=mx+\frac{a}{m}\)
      \( y = 1.x + \frac{3}{2} \)
      \( y = x + \frac{3}{2} \)
      To find the point of contact, we substitute the tangent equation \( y = x +\frac{3}{2}\) into the parabola equation \( y^2 = 6x \), so we get
      \( y^2 = 6x \)
      \( (x + \frac{3}{2})^2 = 6x \)
      \( (2x + 3)^2 = 24x \)
      \( (2x - 3)^2 = 0 \)
      \( x= \frac{3}{2}\) gives \( y = \frac{3}{2} + \frac{3}{2} =3\)
      Hence, the point of contact is \( (\frac{3}{2},3)\).
   2. A tangent to the parabola \(y^2=12x\) makes an angle 45 degree with the straight line \(2y=x+3\). Find its equation and point of contact.
   3. Find the equation of tangents to the parabola \(y^2=12x\) drawn through the points (-1,2). Also find the point of contact and the angle between two tangents.
   4. Find the equation of common tangents to the parabolas \( y^2=4x\) and \( x^2=4y\)
6. Solve the followings
   1. Show that the pair of tangents drawn from the point (-2,3) to the parabola \(y^2=8x\) are at right angle
   2. Prove that the tangents at the extrimities of the latus rectum of a parabola \(y^2=16x\) are at right angles