MHT CET · Maths · Ellipse
Tangent to the ellipse \(\frac{x^{2}}{32}+\frac{y^{2}}{18}=1\) having slope \(-\frac{3}{4}\) meet the coordinate axes in \(A\) and \(B\). Find the area of the \(\Delta A O B\), where \(O\) is the origin.
- A 12 sq unit
- B \(8 \mathrm{sq}\) unit
- C 24 sq unit
- D 32 sq unit
Answer & Solution
Correct Answer
(C) 24 sq unit
Step-by-step Solution
Detailed explanation
Equation of tangent with slope \(-\frac{3}{4}\) is
\(
y=-\frac{3}{4} x+c
\)
According to condition of tengency
\(\therefore\)
\(
\begin{aligned}
c &=\sqrt{32 \times\left(\frac{-3}{4}\right)^{2}+18} \\
&=\sqrt{18+18} \\
&=6 \\
y &=-\frac{3}{4} x+6
\end{aligned}
\)
\(
\Rightarrow 4 y+3 x=24
\)
It meets the coordinate axes in \(A\) and \(B\). \(\therefore A \equiv(8,0)\) and \(B \equiv(0,6)\)
Required area \(=\frac{1}{2} \times 8 \times 6=24\) sq unit
\(
y=-\frac{3}{4} x+c
\)
According to condition of tengency
\(\therefore\)
\(
\begin{aligned}
c &=\sqrt{32 \times\left(\frac{-3}{4}\right)^{2}+18} \\
&=\sqrt{18+18} \\
&=6 \\
y &=-\frac{3}{4} x+6
\end{aligned}
\)
\(
\Rightarrow 4 y+3 x=24
\)
It meets the coordinate axes in \(A\) and \(B\). \(\therefore A \equiv(8,0)\) and \(B \equiv(0,6)\)
Required area \(=\frac{1}{2} \times 8 \times 6=24\) sq unit
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