Question

If the line joining the points $(0,5)$ and $(5,3)$ is a tangent to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$, then:

• A. $a^2<\frac{625}{8},b^2<\frac{25}{2}$
• B. $a^2<\frac{625}{16},b^2>\frac{25}{4}$
• C. $1<a^2<2$
• D. $2<b^2<3$

Answer 1

The correct option is A

$a^2<\frac{625}{8},b^2<\frac{25}{2}$

Step 1: Evaluate the equation of the line passing through the given points

We know that the equation of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\therefore$The equation of the line passing through the points $(0,5)$ and $(5,3)$ will be $\frac{y-5}{3-5}=\frac{x-0}{5-0}\Rightarrow y=-\frac{2}{5}x+5$ $...\left(1\right)$

Step 2: Use the condition for a line to touches the ellipse

We know that the line $y=mx+c$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if $c^2=a^2{m}^2+b^2$

Given curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$ is an ellipse as $\frac{x^2}{{\left(\sqrt{2}a\right)}^2}+\frac{y^2}{{\left(\sqrt{2}b\right)}^2}=1.....\left(2\right)$

It is given that the line joining the points $(0,5)$ and $(5,3)$ is a tangent to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$.

So the line given in $\left(1\right)$ touches ellipse given in $\left(2\right)$

We know that the line $y=mx+c$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if $c^2=a^2{m}^2+b^2$

Therefore, from $\left(1\right)$ and $\left(2\right)$, we have

$\begin{array}{rcl}{\left(5\right)}^2& =& {\left(\sqrt{2}a\right)}^2{\left(\frac{-2}{5}\right)}^2+{\left(\sqrt{2}b\right)}^2\\ & \Rightarrow & 25=\frac{8a^2}{25}+2b^2\\ & \Rightarrow & 625=8a^2+50b^2......\left(3\right)\end{array}$

Step 3: Finding the range of $a^2$ and $b^2$

From $\left(3\right)$, we get

$\begin{array}{rcl}8a^2& =& 625-50b^2.....\left(4\right)\\ 50b^2& =& 625-8a^2......\left(5\right)\end{array}$

L.H.S of $\left(4\right)$ is always positive as $\begin{array}{rcl}8a^2& >& 0,\forall a\in \mathrm{ℝ},a^2>0\end{array}$

$\begin{array}{rcl}\therefore 625-50b^2& >& 0\\ & \Rightarrow & 625>50b^2\\ & \Rightarrow & \frac{25}{2}>b^2\end{array}$

Similarly, the L.H.S of $\left(5\right)$ is always positive as $\begin{array}{rcl}50b^2& >& 0,\forall b\in \mathrm{ℝ},b^2>0\end{array}$

$\begin{array}{rcl}\therefore 625-8a^2& >& 0\\ & \Rightarrow & 625>8a^2\\ & \Rightarrow & \frac{625}{8}>a^2\end{array}$

Therefore, the correct option is A.