Question

The line $4x-7y+10=0$ intersect the parabola, $y^2=4x$ at the points $P$ and $Q$. The coordinates of the point of intersection of the tangents drawn at the point $P$ and $Q$ are

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

Answer 1

The correct option is B $(\frac{5}{2},\frac{7}{2})$

According to the figure,

$ky=\frac{4\left(x+h\right)}{2}ky=2\left(x+h\right)2ky=4x+4h4x-2ky+4h=04x-7y+10=04h=107=2kh=\frac{5}{2}k=\frac{7}{2}$

$pointofintersectionof\tan gentatp$ and $qis\left(\frac{5}{2},\frac{7}{2}\right)$

$Hence,optionBisthecorrectanswer.$