The line y = mx + c touches the parabola y^2 | vedclass

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

Question

(D) The equation of the parabola is $y^2 = 4a(x + a)$.Let $X = x + a$,then $x = X - a$. Substituting this into the equation,we get $y^2 = 4aX$.The con

Vedclass · Accepted Answer

$c = am + \frac{a}{m}$

Vedclass · Answer

(D) The equation of the parabola is $y^2 = 4a(x + a)$.Let $X = x + a$,then $x = X - a$. Substituting this into the equation,we get $y^2 = 4aX$.The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4aX$ is $c' = \frac{a}{m}$,where $c'$ is the intercept on the $Y$-axis.Substituting $x = X - a$ into $y = mx + c$,we get $y = m(X - a) + c = mX + (c - am)$.Comparing this with $y = mX + c'$,we have $c' = c - am$.Equating the two expressions for $c'$,we get $c - am = \frac{a}{m}$.Therefore,$c = am + \frac{a}{m}$.

The line $y = mx + c$ touches the parabola $y^2 = 4a(x + a)$ if...

Similar Questions

If $P\left(\frac{1}{2}, 4\right)$ and $Q$ are the ends of a focal chord of the parabola $y^2=32x$ and $S$ is the focus of the parabola,then $SQ=$

What are the parametric equations of the parabola $(y - 2)^2 = 12(x - 4)$?

The curve described parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represents:

The diameter of the parabola $y^2 = 4ax$ which bisects the chords parallel to $y = mx + \alpha$ is:

Find the equation of the chord of the parabola $y^2 = 6x$ passing through the vertex and the negative end of the latus rectum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module