Question

For the value of $a=\sqrt{2}$ if a tangent is drawn to a suitable conic (Column 1) at the point of contact $\left( -1,1 \right)$, then which of the following options is the correct combination for obtaining its equation?

A. I,(ii),Q
B. III,(i), P
C. II,(ii),Q
D. I,(i), P

Answer 1

To solve this question, we should know the concept of the slope of tangent at a point on the curve. To solve this question, we should first get the general equation of the tangent to different curves given. To do that, we should assume a line $y=mx+c$ and we know that the condition for the line to be a tangent to the curve is that it should have only a single point of intersection with the curve $y=f\left( x \right)$ in column-1. The condition for equal roots of the equation $a{{x}^{2}}+bx+c=0$ is given by${{b}^{2}}=4ac$ and the root of the equation will be $x=\dfrac{-b}{2a}$. So, we should solve both the equations and get the relation between $m\text{ and }c$ and the general equation of tangent. We can also get the parametric point at which the general tangent is drawn. Once we get this, we can match the curves in the column-1 to the tangent and point in column-2 and 3 respectively. By using the point $\left( -1,1 \right)$ and $a=\sqrt{2}$, we can get the required answer.

Complete step by step answer:
To solve this question, we will consider the curves in column-1 and the general tangent equation as $y=mx+c$ and solve further.
Let us consider the first curve ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$
Let the tangent to this curve be $y=mx+c$.
We will substitute the value of $y=mx+c$ in place of y in the equation of the curve. We get
$\begin{aligned} & {{x}^{2}}+{{\left( mx+c \right)}^{2}}={{a}^{2}} \\\ & {{x}^{2}}+{{m}^{2}}{{x}^{2}}+{{c}^{2}}-2cmx={{a}^{2}} \\\ & \left( 1+{{m}^{2}} \right){{x}^{2}}-2cmx+{{c}^{2}}-{{a}^{2}}=0\to \left( 1 \right) \\\ \end{aligned}$
The condition for equal roots of the equation $a{{x}^{2}}+bx+c=0$ is given by${{b}^{2}}=4ac$ and the root of the equation will be $x=\dfrac{-b}{2a}$.
We know that the line $y=mx+c$ is a tangent to the curve and only one solution should be there. So, we can write from the above condition that
$\begin{aligned} & {{\left( -2cm \right)}^{2}}=4\left( {{c}^{2}}-{{a}^{2}} \right)\left( 1+{{m}^{2}} \right) \\\ & 4{{c}^{2}}{{m}^{2}}=4\left( {{c}^{2}}-{{a}^{2}}-{{a}^{2}}{{m}^{2}}+{{c}^{2}}{{m}^{2}} \right) \\\ & 4{{c}^{2}}{{m}^{2}}=4{{c}^{2}}-4{{a}^{2}}-4{{a}^{2}}{{m}^{2}}+4{{c}^{2}}{{m}^{2}} \\\ & 4{{c}^{2}}=4{{a}^{2}}+4{{a}^{2}}{{m}^{2}} \\\ & {{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right) \\\ & c=\pm a\sqrt{1+{{m}^{2}}} \\\ \end{aligned}$
Using this, we get the tangent as $y=mx\pm a\sqrt{1+{{m}^{2}}}\to \left( 2 \right)$
We have the point as $\left( -1,1 \right)$ and the value of a as $a=\sqrt{2}$. Using them in the above equation, we get
$\begin{aligned} & 1=m\left( -1 \right)\pm \sqrt{2}\sqrt{1+{{m}^{2}}} \\\ & 1+m=\pm \sqrt{2}\sqrt{1+{{m}^{2}}} \\\ \end{aligned}$
Squaring on both sides, we get

& {{\left( 1+m \right)}^{2}}={{\left( \pm \sqrt{2}\sqrt{1+{{m}^{2}}} \right)}^{2}} \\\ & {{m}^{2}}+1+2m=2\left( 1+{{m}^{2}} \right) \\\ & {{m}^{2}}+2m+1-2-2{{m}^{2}}=0 \\\ & {{m}^{2}}-2m+1=0 \\\ & {{\left( m-1 \right)}^{2}}=0 \\\ & m=1 \\\ \end{aligned}$$ So, we got the value of m as $m=1$ Now, we should use the values of $m=1$ and $a=\sqrt{2}$ in the column-3 if we get the point $\left( -1,1 \right)$. If we don’t get any correct answer, we should repeat the whole process for the second curve. Checking the row P, we get $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)=\left( \dfrac{\sqrt{2}}{{{1}^{2}}},\dfrac{2\sqrt{2}}{1} \right)=\left( \sqrt{2},2\sqrt{2} \right)$ We can infer that row-P is not matching. Checking the row Q, we get $\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)=\left( \dfrac{-\sqrt{2}\times 1}{\sqrt{1+{{1}^{2}}}},\dfrac{\sqrt{2}}{\sqrt{1+{{1}^{2}}}} \right)=\left( \dfrac{-\sqrt{2}}{\sqrt{2}},\dfrac{\sqrt{2}}{\sqrt{2}} \right)=\left( -1,1 \right)$ So, row-Q is matching with the values. The equation of tangent is $y=x\pm \sqrt{2}\sqrt{1+{{1}^{2}}}=x\pm 2$ In our case, when $x=-1,y=1$ So, we get $y=x+2$ as the equation of the tangent. $y=x+2$ is the equation of tangent we get by substituting $a=\sqrt{2}$ and $m=1$ in the equation $y=mx+a\sqrt{{{m}^{2}}+1}$. So, we are getting a correct match when the curve is I in column-1 and tangent (ii) in column-2 and the point of contact is Q in column-3. The given question is a single correct answer and the answer is option-A **So, the correct answer is “Option A”.** **Note:** The general equations of tangent to different standard curves are important and worthy enough to be remembered. The curve that we got as an answer is a circle and the general equation of the tangent of a circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is given by $y=mx\pm a\sqrt{1+{{m}^{2}}}$. Likewise, by following the procedure, we can find the general equations of tangents for all the curves given in the table. One trick to get to the answer in this question quickly is by substituting the value of $a=\sqrt{2}$ and $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -1,1 \right)$ in the curves directly. We can infer from this that only the first curve satisfies the given constraints. After that, our procedure gets easier.