MHT CET · Maths · Ellipse
The equation of tangent to the curve \(9 x^{2}+16 y^{2}=144\) which makes equal intercepts with coordinate axes, is
- A \(x+y=5\)
- B \(x+y=16\)
- C \(x+y=15\)
- D None of these
Answer & Solution
Correct Answer
(A) \(x+y=5\)
Step-by-step Solution
Detailed explanation
Given curve is \(9 x^{2}+16 y^{2}=144\)
\(
\Rightarrow \quad \frac{x^{2}}{16}+\frac{y^{2}}{9}=1
\)
Let the equation of tangent is \(x+y=k\)
\(\Rightarrow \quad y=-x+k\)
It is a tangent to the given curve if
\(
\begin{aligned}
& c^{2}=a^{2} m^{2}+b^{2} \\
\Rightarrow \quad & k^{2}=16(-1)^{2}+9
\end{aligned}
\)
\(\Rightarrow \quad k^{2}=25\)
\(
\Rightarrow \quad k=\pm 5
\)
\(\therefore\) Required equations of tangent are
\(
x+y=\pm 5
\)
\(
\Rightarrow \quad \frac{x^{2}}{16}+\frac{y^{2}}{9}=1
\)
Let the equation of tangent is \(x+y=k\)
\(\Rightarrow \quad y=-x+k\)
It is a tangent to the given curve if
\(
\begin{aligned}
& c^{2}=a^{2} m^{2}+b^{2} \\
\Rightarrow \quad & k^{2}=16(-1)^{2}+9
\end{aligned}
\)
\(\Rightarrow \quad k^{2}=25\)
\(
\Rightarrow \quad k=\pm 5
\)
\(\therefore\) Required equations of tangent are
\(
x+y=\pm 5
\)
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