Question

If a tangent to a suitable conic (Column I) is found to be $y=x+8$ and its point of contact is $(8,16)$, then which of the following options is the only CORRECT combination?

• A. $\left(III\right)\left(ii\right)\left(Q\right)$
• B. $\left(III\right)\left(i\right)\left(P\right)$
• C. $\left(II\right)\left(iv\right)\left(R\right)$
• D. $\left(I\right)\left(ii\right)\left(Q\right)$

Answer 1

Answer: (B)

$\left(III\right)\left(i\right)\left(P\right)$

Point of contact : $(8,16)$

Equation of tangent $\Rightarrow y=x+8$

$\therefore$ Slope $\left(m\right)=1$ .

i.e. Matching with $\left(i\right)$ of column 2

$my=m^{2}x+a$

$y=x+8$

Hence, $m=1$ and $a=8$

Matching with $\left(P\right)$ of column 3 : $(\frac{a}{m^2},\frac{2a}{m})$

$\Rightarrow (\frac{8}{1},\frac{2\times 8}{1})=(8,16)$

Matching with column 1

1) $x^2+y^2-a^2=0\Rightarrow 8^2+{16}^2-8^2\ne 0$

2) $x^2+a^2{y}^2-a^2=0\Rightarrow 8^2+(8^2\times {16}^2)-8^2\ne 0$

3) $y^2=4ax\Rightarrow {16}^2-4\left(8\right)\left(8\right)=0$

Hence, equation of curve $\Rightarrow y^2=4ax$

Point of contact : $(\frac{a}{m^2},\frac{2a}{m})$

Tangent $\Rightarrow my=m^{2}x+a$

Hence, option B is correct.