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Question: If the tangents on the ellipse \[4{{x}^{2}}+{{y}^{2}}=8\] at the point \(\left( 1,2 \right)\) and \(...

If the tangents on the ellipse 4x2+y2=84{{x}^{2}}+{{y}^{2}}=8 at the point (1,2)\left( 1,2 \right) and (a,b)\left( a,b \right) are perpendicular to each other, then a2{{a}^{2}} is equal to:
A. 6417\dfrac{64}{17}
B. 217\dfrac{2}{17}
C. 12817\dfrac{128}{17}
D. 417\dfrac{4}{17}

Explanation

Solution

First we will start by converting the given equation of ellipse into its standard form and then we will find the equation of tangents at the given point (1,2)\left( 1,2 \right) and (a,b)\left( a,b \right) using the formula xx1a2+yy1b2=1\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1 ; once we find both the tangents, we will then apply the theorem of perpendicular lines and find the variables.

Complete step by step answer:
We will start by converting the given equation of ellipse into its standard form that is: x2a2+y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1
Now, we have the equation of ellipse given in the question as follows: 4x2+y2=8 ........... Equation 1.4{{x}^{2}}+{{y}^{2}}=8\text{ }...........\text{ Equation 1}\text{.}
Now we will divide the whole equation by 8 to convert it into its standard form:

& \Rightarrow ~\dfrac{4{{x}^{2}}}{8}+\dfrac{{{y}^{2}}}{8}=\dfrac{8}{8} \\\ & \Rightarrow \dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{8}=1 \\\ & \Rightarrow \dfrac{{{x}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}=1 \\\ \end{aligned}$$ Now, we know that the equation of tangent to the ellipse $$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$$ at $$({{x}_{1}},{{y}_{1}})$$ is $$\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$$ Therefore, equation of tangent to the ellipse $$\dfrac{{{x}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}=1$$ at $\left( 1,2 \right)$ is : $$\begin{aligned} & \dfrac{x\left( 1 \right)}{{{\left( \sqrt{2} \right)}^{2}}}+\dfrac{y\left( 2 \right)}{{{\left( 2\sqrt{2} \right)}^{2}}}=1\Rightarrow \dfrac{x}{2}+\dfrac{y}{4}=1 \\\ & \Rightarrow \dfrac{2x+y}{4}=1\text{ }\Rightarrow 2x+y=4.........\text{ Equation 2}\text{.} \\\ \end{aligned}$$ Similarly, the equation of tangent at $\left( a,b \right)$ will be : $$\begin{aligned} & \dfrac{x\left( a \right)}{{{\left( \sqrt{2} \right)}^{2}}}+\dfrac{y\left( b \right)}{{{\left( 2\sqrt{2} \right)}^{2}}}=1\Rightarrow \dfrac{ax}{2}+\dfrac{by}{8}=1 \\\ & \Rightarrow \dfrac{4ax+by}{8}=1\Rightarrow 4ax+by=8\text{ }.........\text{ Equation 3}\text{.} \\\ \end{aligned}$$ So according to these conditions, we will have the following diagram: ![](https://www.vedantu.com/question-sets/0c580681-8584-4baf-9524-1c384c0a57bb1752497293242882434.png) Now, it is given in the question that both the tangents at $\left( 1,2 \right)$ and $\left( a,b \right)$ are perpendicular to each other, so we will apply the following theorem: If lines $${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\ \,\text{and}\ \,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$$ are perpendicular, then $$\left( \dfrac{-{{a}_{1}}}{{{b}_{1}}} \right)\times \left( -\dfrac{{{a}_{2}}}{{{b}_{2}}} \right)=-1$$ , we have with us $$2x+y=4$$and $$4ax+by=8$$ , which are perpendicular to each other : Therefore: $$\left( \dfrac{-2}{1} \right)\times \left( \dfrac{-4a}{b} \right)=-1\Rightarrow b=-8a$$ Now, on squaring both sides we will get ${{b}^{2}}=64{{a}^{2}}\text{ }........\text{Equation 4}\text{.}$ Now it is given in the question that $\left( a,b \right)$ lies on the given ellipse therefore we will put these points in the equation of ellipse as seen in Equation 1 above: $$\begin{aligned} & 4{{x}^{2}}+{{y}^{2}}=8 \\\ & \Rightarrow 4{{\left( a \right)}^{2}}+{{\left( b \right)}^{2}}=8 \\\ & \Rightarrow 4{{a}^{2}}+{{b}^{2}}=8 \\\ \end{aligned}$$ We will now put the value of ${{b}^{2}}$ from equation 4 in $$4{{a}^{2}}+{{b}^{2}}=8$$: $$4{{a}^{2}}+{{b}^{2}}=8\Rightarrow 4{{a}^{2}}+64{{a}^{2}}=8$$ ; Taking out 4 common from the equation : $$\begin{aligned} & \Rightarrow 4{{a}^{2}}+64{{a}^{2}}=8 \\\ & \Rightarrow 4\left( {{a}^{2}}+16{{a}^{2}} \right)=8 \\\ & \Rightarrow {{a}^{2}}+16{{a}^{2}}=2 \\\ & \Rightarrow 17{{a}^{2}}=2 \\\ & \Rightarrow {{a}^{2}}=\dfrac{2}{17} \\\ \end{aligned}$$ **Therefore the answer is option D.** **Note:** Remember that mistakes can be made in applying the perpendicular lines theorem as there are three negative signs involved. We will get the negative value of $b$ , but we will square it as we require the square of $a$ according to the question.