Question

Column 1: Information about curves
Column 2: Type of curves
Column - 3: eccentricity of the curves given in column - 1
$\begin{array}{cccccc}& Column-1& & Column-2& & Column-3\\ \left(I\right)& \text{The curve such that products}& \left(i\right)& \text{Circle}& \left(P\right)& e=1\\ & \text{of the distances of any of its}& & & & \\ & \text{tangent from two given points}& & & & \\ & \text{is constant can be}& & & & \\ \left(II\right)& \text{A curve for which the length}& \left(ii\right)& \text{Parabola}& \left(Q\right)& e<1\\ & \text{of the subnormal at any of it's}& & & & \\ & \text{point is equal to 2 and the curve}& & & & \\ & \text{passes through (1,2) can be}& & & & \\ \left(III\right)& \text{A curve passes through (1,4) and}& \left(iii\right)& \text{Ellipse}& \left(R\right)& e=0\\ & \text{is such that the segment joining}& & & & \\ & \text{any point P on the curve and the}& & & & \\ & \text{point of intersection of the normal}& & & & \\ & \text{at P with the x-axis is bisected by}& & & & \\ & \text{the y-axis. the curve can be}& & & & \\ \left(IV\right)& \text{A curve passes through (1,2) is}& \left(iv\right)& \text{Hyperbola}& \left(S\right)& e>1\\ & \text{such that the length of the}& & & & \\ & \text{normal at any of its point}& & & & \\ & \text{is equal to 2}.& & & & \end{array}$
Which of the following options is the only correct combination?

• A. (I)(i)(P)
• B. (I)(i)(R)
• C. (I)(iii)(Q)
• D. (I)(iv)(Q)

Answer 1

The correct option is C (I)(iii)(Q)

(C)
(I) It is a very important property of ellipse and hyperbola $p_1{p}_2=b^2$

(II)$y\left(\frac{dy}{dx}\right)=2\Rightarrow \frac{y^2}{2}=2x+c$
At $x=1,y=2\Rightarrow c=0$
$\Rightarrow y^2=4x\Rightarrow$ which is a parabola.

(III) Equation of normal at $P$
$(Y-y)=\frac{-1}{m}(X-x)$
$Y=0,X=x+my$
$X=0,Y=y-\frac{x}{m}$
Hence, $x+my+x=0\Rightarrow 2x+y\left(\frac{dy}{dx}\right)=0$
$\left(2x\right)dx+\left(y\right)dy=0$
$x^2+\frac{y^2}{2}=C$
$\Rightarrow 1+8=C$ [passes through (1, 4)]
$\therefore x^2+\frac{y^2}{2}=9\Rightarrow \frac{x^2}{9}+\frac{y^2}{18}=1\Rightarrow$ ellipse

(IV) length of normal
$(x+my-x{)}^2+y^2=4$
$\Rightarrow m^2=\frac{4-y^2}{y^2}\Rightarrow \frac{dy}{dx}=\frac{\sqrt{4-y^2}}{y}\Rightarrow -\sqrt{4-y^2}=x+c$
$x=1,y=4\Rightarrow c=-1$
$\therefore (x-1{)}^2+y^2=4\Rightarrow$ circle.