Question

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the centre $S$ of the circle $x^2+y^2–4x–16y+64=0.$ Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then

• A. $SP=2\sqrt{5}$
• B. $SQ:QP=(\sqrt{5}+1):2$
• C. the $x$-intercept of the normal to the parabola at $P$ is $6$
• D. the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Answer 1

Answer: (A), (C), (D)

the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Equation of the circle:
$x^2+y^2–4x–16y+64=0\Rightarrow (x-2{)}^2+(y-8{)}^2=(2{)}^2$


$PS$ will be minimum when $PS$ is normal to the parabola.
Slope of the tangent of the parabola at $P(x_1,y_1)$ is $\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y_1}$
Slope of the normal at $P(x_1,y_1),$
$m_n=-\frac{y_1}{2}=\frac{y_1-8}{x_1-2}\cdots \left(i\right)\text{Also,}{x}_1=\frac{y_1^2}{4}\cdots \left(ii\right)\text{From}\left(i\right)\&\left(ii\right),-y_1(\frac{y_1^2}{4}-2)=2y_1-16\Rightarrow y_1^3=64\Rightarrow y_1=4$
$\therefore$ Coordinates of $P\text{is}(4,4)$

Option (1):
$SP=\sqrt{(4-2{)}^2+(4-8{)}^2}=2\sqrt{5}$

Option (2):
$SQ=2\Rightarrow QP=SP-SQ=2\sqrt{5}-2\therefore SQ:QP=2:(2\sqrt{5}-2)=1:(\sqrt{5}-1)=(\sqrt{5}-1):4$

Option (3):
Slope of the normal at $P$ is
$m_n=-\frac{y_1}{2}=-2$
Therefore, equation of the normal to the parabola at $P$ is $\frac{y-4}{x-4}=-2\Rightarrow y+2x=12$

Option (4):
Slope of the tangent at $Q$ is
$m_t{|}_Q=-\frac{1}{m_n}=\frac{1}{2}$