The point of contact of the tangent 18x - 6y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the point of contact be $(h, k)$.Since the point $(h, k)$ lies on the parabola $y^2 = 2x$,we have $k^2 = 2h$.The equation of the tangent to th

Vedclass · Accepted Answer

$\left( \frac{1}{18}, \frac{1}{3} \right)$

Vedclass · Answer

(D) Let the point of contact be $(h, k)$.Since the point $(h, k)$ lies on the parabola $y^2 = 2x$,we have $k^2 = 2h$.The equation of the tangent to the parabola $y^2 = 4ax$ at $(h, k)$ is $ky = 2a(x + h)$.Here,$4a = 2$,so $a = \frac{1}{2}$.Thus,the tangent equation is $ky = 1(x + h)$,or $x - ky + h = 0$.Comparing this with the given tangent $18x - 6y + 1 = 0$,we get:$\frac{1}{18} = \frac{-k}{-6} = \frac{h}{1}$.From $\frac{1}{18} = \frac{k}{6}$,we get $k = \frac{6}{18} = \frac{1}{3}$.From $\frac{1}{18} = h$,we get $h = \frac{1}{18}$.Thus,the point of contact is $\left( \frac{1}{18}, \frac{1}{3} \right)$.

The point of contact of the tangent $18x - 6y + 1 = 0$ to the parabola $y^2 = 2x$ is

Similar Questions

If $x + y = k$ is a normal to the parabola ${y^2} = 12x$,then $k$ is

Let $O$ be the origin and $A$ be a point on the curve $y^2=4x$. Then the locus of the midpoint of $OA$ is:

The equations of the tangents to the parabola $y^2 = 4ax$ at the ends of its latus rectum are-

Find the length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes an angle of $45^{\circ}$ with the $x$-axis.

What are the equations of the tangents at the endpoints of the latus rectum of the parabola $y^2 = 4ax$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module