Question

$y^2=4x$ is a curve and $P$, $Q$ and $R$ are three points on it, where $P=(1,2)$, $Q=(\frac{1}{4},1)$ and the tangent to the curve at $R$ is parallel to the chord $PQ$ of the curve, then coordinates of $R$ are

• A. $(\frac{5}{8},\sqrt{\frac{5}{2}})$
• B. $(\frac{9}{16},\frac{3}{2})$
• C. $(\frac{5}{8},-\sqrt{\frac{5}{2}})$
• D. $(\frac{9}{16},\frac{-3}{2})$

Answer 1

The correct option is B $(\frac{9}{16},\frac{3}{2})$

$y^2=4xP(1,2),Q(\frac{1}{4},1)$

Slope of $PQ$$=\frac{1-2}{\frac{1}{4}-1}=\frac{4}{3}$

Slope of tangent $=\frac{4}{3}$

Equation of tangent in slope form is $y=mx+\frac{a}{m}$

$a=1$

$y=\frac{4}{3}x+\frac{1}{\frac{4}{3}}y=\frac{4}{3}x+\frac{3}{4}12y=16x+\mathrm{9....}\left(i\right)16x=12y-94x=\frac{12y-9}{4}$

Substituting in the equation of parabola

$y^2=\frac{12y-9}{4}4y^2=12y-94y^2-12y+9=0{(2y-3)}^2=0\Rightarrow y=\frac{3}{2}{\left(\frac{3}{2}\right)}^2=4x\Rightarrow x=\frac{9}{16}$

So the point $R$ is $(\frac{9}{16},\frac{3}{2})$

Option $\left(B\right)$ is correct.