Question

The line $4x-7y+10=0$ intersects the parabola , $y^2=4ax$ at the points A & B . The co-ordinates of the point of intersection of the tangents drawn at the points A & B are :

• A. $\left(\frac{7}{2},\frac{5}{2}\right)$
• B. $\left(\frac{5}{3},\frac{8}{3}\right)$
• C. $\left(\frac{5}{2},\frac{7}{2}\right)$
• D. $\left(-\frac{7}{2},-\frac{5}{2}\right)$

Answer 1

The correct option is B $\left(\frac{5}{3},\frac{8}{3}\right)$

Given:Line $4x=7y-10$

intersects the parabola $y^2=4x$

$\Rightarrow y^2=7y-10$

or $y^2-7y+10=0$

$\Rightarrow (y-5)(y-2)=0$

$\therefore y=5,2$

Put $y=5$ in $y^2=4x$ we get

$4x=25$

or $x=\frac{25}{4}$

Again Put $y=2$ in $y^2=4x$ we get

$4x=4$

or $x=1$

$\therefore A(1,2)$ and $B(\frac{25}{4},5)$

Tangent of $y^2=4x$ is $T=0$ is $y{y}_1=2(x+x_1)$

At $A(1,2)$ we have $2y=2(x+1)$

or $y=x+1$ or $x-y+1=0$ is the equation of the tangent at the point $A$

At $B(\frac{25}{4},5)$ we have $5y=2(x+\frac{25}{4})$

or $20y=8x+50$ or $4x-10y-25=0$ is the equation of the tangent at the point $B$

$\Rightarrow 10x+10=4x+25$

$\Rightarrow 10x-4x=25-10$

$\Rightarrow 6x=15$

or $x=\frac{5}{3}$

Substituting $x=\frac{5}{3}$ in $x-y+1=0$ we get

$\frac{5}{3}-y+1=0$

$\Rightarrow -y+\frac{8}{3}=0$

$\therefore y=\frac{8}{3}$

$\therefore$ tangent intersect at the point $(\frac{5}{3},\frac{8}{3})$