Question

Column 1, 2, 3 contains conics, equation of tangents, and points of contact respectively.$$$$

Column I	Column II	Column III
(I)${{x}^{2}}+{{y}^{2}}={{a}^{2}}$	(i)$my={{m}^{2}}x+a$	(P) $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$
(II)${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$	(ii)$y=mx+a\sqrt{{{m}^{2}}+1}$	(Q) $\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$
(III)${{y}^{2}}=4ax$	(iii)$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}$	(R)$\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$
(IV)${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$	(iv)$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}$	(S)$\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)$

Which of the following options is the only CORRECT combination? $A. (II) (iii) (S)$
B. (I) (ii) (R) $C. (III) (i) (P)$
D. (IV)(i)(S) $$$$

Answer 1

We assume the equation tangent as $y=mx+c$ where $m$ is the slope and $c$ is the $y-$intercept. We put $y$ in the equation of each conic and use the condition on discriminant $D={{b}^{2}}-4ac=0$ for rational roots as coordinate point of contact. We find $c$ and the the point of contact using quadratic formula.$$$$

Complete step-by-step solution:
We know that the quadratic equation of the form $a{{x}^{2}}+bx+c=0$ where $a\ne 0,b,c\in R$ will have rational roots when the discriminant $D={{b}^{2}}-4ac=0$ . The roots of the equations are given by the quadratic formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Let us assume that the tangent of all the conics in the column-I of the question in slope-intercept form is
$y=mx+c....\left( 1 \right)$
Here $m$is the slope and $c$is the $y-$intercept of the tangent. Now let us find the intercepts of the conics in column-I.$$$$
(I) The given equation of conic is
${{x}^{2}}+{{y}^{2}}={{a}^{2}}$
Let us put $y=mx+c$ in the above equation and have,

& {{x}^{2}}+{{\left( mx+c \right)}^{2}}={{a}^{2}} \\\ & \Rightarrow {{x}^{2}}+{{m}^{2}}{{x}^{2}}+{{c}^{2}}-{{a}^{2}}+2cmx=0 \\\ & \Rightarrow \left( 1+{{m}^{2}} \right){{x}^{2}}+2cmx+{{c}^{2}}-{{a}^{2}}=0 \\\ \end{aligned}$$ The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we find, $$\begin{aligned} & D={{\left( 2cm \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\\ & \Rightarrow 4{{c}^{2}}{{m}^{2}}-4\left( {{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}} \right)=0 \\\ & \Rightarrow 4\left( {{c}^{2}}{{m}^{2}}-{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}} \right)=0 \\\ & \Rightarrow {{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}} \\\ & \Rightarrow c=\pm a\sqrt{1+{{m}^{2}}} \\\ \end{aligned}$$ So the equation of the tangent is $$y=mx\pm a\sqrt{{{m}^{2}}+1}$$ The coordinates of the point of contact of the conic are roots which we find by putting $c=a\sqrt{1+{{m}^{2}}},{{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}}$ from above calculation of intercept above. We have the $x-$coordinate of point of contact as, $$\begin{aligned} & x=\dfrac{-2cm \pm 4\left( -{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}{2\left( 1+{{m}^{2}} \right)} \\\ & =\dfrac{-2\left( a\sqrt{1+{{m}^{2}}} \right)m \pm \sqrt{4\left( -{{a}^{2}}-{{a}^{2}}{{m}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}}{2\left( 1+{{m}^{2}} \right)} \\\ & =\dfrac{-am}{\sqrt{1+{{m}^{2}}}} \\\ \end{aligned}$$ Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as, $$y=m\left( \dfrac{-am}{\sqrt{1+{{m}^{2}}}} \right)+a\sqrt{1+{{m}^{2}}}=\dfrac{a}{\sqrt{1+{{m}^{2}}}}$$ So the point of contact is $\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (I)(ii)(Q). (II) The given equation of conic is $${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$$ Let us put $y=mx+c$ in the above equation and have, $$\begin{aligned} & {{x}^{2}}+{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\\ & \Rightarrow {{x}^{2}}+{{a}^{2}}{{m}^{2}}{{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\\ & \Rightarrow \left( 1+{{a}^{2}}{{m}^{2}} \right){{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\\ \end{aligned}$$ The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find , $$\begin{aligned} & D={{\left( 2{{a}^{2}}cm \right)}^{2}}-4\left( 1+{{a}^{2}}{{m}^{2}} \right)\left( {{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\\ & \Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}-{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\\ & \Rightarrow 4\left( -{{a}^{2}}{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\\ & \Rightarrow {{a}^{2}}+{{a}^{4}}{{m}^{2}}={{a}^{2}}{{c}^{2}} \\\ & \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}+1 \\\ & \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}+1} \\\ \end{aligned}$$ So the equation of the tangent is $$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}$$ The coordinates of the point of contact of the conic are roots which we find by putting $${{c}^{2}}={{a}^{2}}{{m}^{2}}+1,{{c}^{2}}=\sqrt{{{a}^{2}}{{m}^{2}}+1}$$ from above calculation of intercept. We have the $x-$coordinate of point of contact as, $$\begin{aligned} & x=\dfrac{-2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}{{c}^{2}} \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\\ & =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}+1 \right) \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\\ & =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\\ & =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \\\ \end{aligned}$$ Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as, $$y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}+1}=\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}}$$ So the point of contact is $\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (II)(iv)(R)$$$$ (III) The given equation of conic is $${{y}^{2}}=4ax$$ Let us put $y=mx+c$ in the above equation and have, $$\begin{aligned} & {{\left( mx+c \right)}^{2}}=4ax \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+{{c}^{2}}+2cmx-4ax=0 \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+\left( 2cm-4a \right)x+{{c}^{2}}=0 \\\ \end{aligned}$$ The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we have, $$\begin{aligned} & D={{\left( 2cm-4a \right)}^{2}}-4{{m}^{2}}{{c}^{2}}=0 \\\ & \Rightarrow 4{{c}^{2}}{{m}^{2}}+16{{a}^{2}}-16cma-4{{c}^{2}}{{m}^{2}}=0 \\\ & \Rightarrow 16{{a}^{2}}-16cma=0 \\\ & \Rightarrow c=\dfrac{a}{m} \\\ \end{aligned}$$ So the equation of the tangent is $$\begin{aligned} & y=mx+\dfrac{a}{m} \\\ & \Rightarrow my={{m}^{2}}x+a \\\ \end{aligned}$$ The coordinates of the point of contact of the conic are roots which we find by putting $c=\dfrac{a}{m},{{c}^{2}}=\dfrac{{{a}^{2}}}{{{m}^{2}}}$ from calculation of intercept above. We have the $x-$coordinate of point of contact as, $$\begin{aligned} & x=\dfrac{-\left( 2cm-4a \right)\pm\sqrt{16{{a}^{2}}-16cma}}{2{{m}^{2}}} \\\ & =\dfrac{-2\left( \dfrac{a}{m} \right)m+4a \pm \sqrt{16{{a}^{2}}-16\left( \dfrac{a}{m} \right)ma}}{2{{m}^{2}}} \\\ & =\dfrac{a}{{{m}^{2}}} \\\ \end{aligned}$$ Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as, $$y=m\left( \dfrac{a}{{{m}^{2}}} \right)+\dfrac{a}{m}=\dfrac{2a}{m}$$ So the point of contact is $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$. We match the columns with the obtained result and find the matching as (III)(i)(P).$$$$ (IV) The given equation of conic is $${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$$ Let us put $y=mx+c$ in the above equation and have, $$\begin{aligned} & {{x}^{2}}-{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\\ & \Rightarrow {{x}^{2}}-{{a}^{2}}{{m}^{2}}{{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\\ & \Rightarrow \left( 1-{{a}^{2}}{{m}^{2}} \right){{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\\ \end{aligned}$$ The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find, $$\begin{aligned} & D={{\left( -2{{a}^{2}}cm \right)}^{2}}-4\left( 1-{{a}^{2}}{{m}^{2}} \right)\left( -{{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\\ &\Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\\ & \Rightarrow 4\left( {{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\\ & \Rightarrow {{a}^{2}}-{{a}^{4}}{{m}^{2}}=-{{a}^{2}}{{c}^{2}} \\\ & \Rightarrow 1-{{a}^{2}}{{m}^{2}}=-{{c}^{2}} \\\ & \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}-1 \\\ & \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}-1} \\\ \end{aligned}$$ So the equation of the tangent is $$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}$$ The coordinates of the point of contact of the conic are roots which we find by putting $${{c}^{2}}={{a}^{2}}{{m}^{2}}-1,c=\sqrt{{{a}^{2}}{{m}^{2}}-1}$$ from above calculation of intercept. We have the $x-$coordinate of point of contact as, $$\begin{aligned} & x=\dfrac{2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}{{c}^{2}} \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\\ & =\dfrac{2{{a}^{2}}m\left( \sqrt{{{a}^{2}}{{m}^{2}}-1} \right)\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}-1 \right) \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\\ & =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}-1}}{2\left( {{a}^{2}}{{m}^{2}}-1 \right)} \\\ & =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \\\ \end{aligned}$$ Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as, $$y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}-1}=\dfrac{-1}{\sqrt{1+{{m}^{2}}}}$$ So the point of contact is $$\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)$$. We match the columns with the obtained result and find the matching as (IV)(iii)(S). **So the correct matches are (IV)(iii)(S), (III)(i)(P), (II)(iv)(R), (I)(ii)(Q).We are asked to choose the only CORRECT combination in the options which is the option C.** **Note:** We note that the equation ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is an equation of circle in centre-radius from, ${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of ellipse in axis form, ${{y}^{2}}=4ax$ is an equation of parabola in vertex from and ${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of hyperbola in axis from. We can directly find tangents and point of contact if we remember the formulae.