Question

The point of intersection of tangents at the ends of the latus-rectum of the parabola $x^2+y^2-18x+16y+120=0$ is equal to :

• A. (1,0)
• B. $x^2+4x+2y=0$
• C. $2y+3=0$
• D. $3y=2$

Answer 1

Answer: (B)

Equation to the tangent at $(x_1{y}_1)$ on the parabola $y^2=4ax$ is $y{y}_1=2a(x+x_1)$ $\therefore$ In this case, $a=1$ The co-ordinates at the ends of the latus rectum of the parabola $y^2=4x$ are L (1, 2) and $L_1(1,-2)$ Equation of tangent at L and $L_1$ are $2y=(2x+1)$ and $-2y=2(x+1)$ which gives $x=-1,y=0$
thus the required point of intersection is $(-1,0)$