The point of contact of the tangent to the pa | vedclass

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

Question

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.Here, $4a = 9$, so $a = \frac{9}{4}$.The tangent

Vedclass · Accepted Answer

$\left(\frac{4}{9}, 2\right)$

Vedclass · Answer

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.Here, $4a = 9$, so $a = \frac{9}{4}$.The tangent passes through $(4, 10)$, so $10t = 4 + \frac{9}{4}t^2$.Multiplying by $4$, we get $40t = 16 + 9t^2$, or $9t^2 - 40t + 16 = 0$.Factoring the quadratic: $(9t - 4)(t - 4) = 0$, which gives $t = 4$ or $t = \frac{4}{9}$.The slope of the tangent is $m = \frac{1}{t} = 	an 	heta$.Given $	an 	heta > 2$, we have $\frac{1}{t} > 2$, which implies $t Thus, $t = \frac{4}{9}$ is the valid parameter.The point of contact is $(at^2, 2at) = \left(\frac{9}{4} 	imes \left(\frac{4}{9}ight)^2, 2 	imes \frac{9}{4} 	imes \frac{4}{9}ight) = \left(\frac{4}{9}, 2ight)$.

The point of contact of the tangent to the parabola $y^2=9x$ which passes through the point $(4, 10)$ and makes an angle $\theta$ with the positive side of the axis of the parabola where $\tan \theta > 2$, is

Similar Questions

If the line $x = my + k$ touches the parabola $x^2 = 4ay$,then $k = $

The equation of the tangent to the parabola $y^{2}=4x$ inclined at an angle of $\frac{\pi}{4}$ to the positive direction of $x$-axis is:

From the point $(-1, 2)$,tangent lines are drawn to the parabola $y^2 = 4x$. Find the equation of the chord of contact.

What is the equation of the tangent to the parabola $y^2 = 4x$ at the point $(1, 2)$?

The parabola $x^2=4ay$ makes an intercept of length $\sqrt{40}$ units on the line $y=2x+1$. Find the value of $4a$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module