Question

The tangents to the curve $y=(x-2{)}^2-1$ at its points of intersection with the line $x-y=3$, intersect at the point :

• A. $(\frac{5}{2},-1)$
• B. $(-\frac{5}{2},-1)$
• C. $(\frac{5}{2},1)$
• D. $(-\frac{5}{2},1)$

Answer 1

The correct option is A $(\frac{5}{2},-1)$

$x-y-3=0\cdots \left(1\right)$
will be chord of contact of the parabola
$y=x^2-4x+3$
Let the required point is $P(x_1,y_1)$, then chord of contact for point $P$ is
$\frac{y+y_1}{2}=x{x}_1-4\frac{(x+x_1)}{2}+3(2x_1-4)x-y+(-4x_1-y_1+6)=0\cdots \left(2\right)$

Comparing equation $\left(1\right)$ and $\left(2\right)$
$2x_1-4=1,-4x_1-y_1+6=-3x_1=\frac{5}{2},y_1=-1$
So, required point is $(\frac{5}{2},-1)$