#### Tangent to Parabola at \( (x_1, y_1) \)

#### Condition of Tangenncy to Parabola

#### Normal to Parabola at \( (x_1, y_1) \)

#### Tangent and Normal to Parabola in Parametric Form

#### Exercise

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

- \( x^2=12y\) at (-6,3)
- \( x^2=4y\) at the point whose abscissa is 6
- \( y^2=16ax\) at the point whose ordinate is -4a
- \( y^2=12x\) at each end of the latus rectum

- \( y^2=8x\) at (2,-4)
- Solve the followings
- Prove that the line \(3x+4y+6=0\) is tangent to the parabola \(2y^2=9x\) and find its point of contact

- Prove that the line \(lx+my+n=0\) touches the parabola \(y^2=4ax\) if \(ln =am^2\)

- For what value of a, will the straight line \(y=2x+3\) touch the parabola \(y^2=4ax\)

- If the line \(2x+4y=3\) touches the parabola \(y^2=4ax\) find the length of the latus rectum

- Prove that the line \(3x+4y+6=0\) is tangent to the parabola \(2y^2=9x\) and find its point of contact
- Solve the followings
- 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. - 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 \(3x+2y=8\)

- Find the equation of tangent to the parabola \(y^2=4x\)
- Solve the followings
- 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 - 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 \( (18, -12) \)

- Show that tangent to the parabola \(y^2=4x\) and \(x^2=4y\) at (1,2) and (-2,1) respectively are at right angles.
- Solve the followings
- 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 \( (\frac{3}{2},3)\). - 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.
- 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.
- Find the equation of common tangents to the parabolas \( y^2=4x\) and \( x^2=4y\)

- 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.
- Solve the followings
- Show that the pair of tangents drawn from the point (-2,3) to the parabola \(y^2=8x\) are at right angle
- Prove that the tangents at the extrimities of the latus rectum of a parabola \(y^2=16x\) are at right angles

nice

ReplyDelete