MHT CET · Maths · Parabola
The equation of common tangent to the circle \(x^{2}+y^{2}=2\) and parabola \(y^{2}=8 x\) is
- A \(y=x+1\)
- B \(y=x+2\)
- C \(y=x-2\)
- D \(y=-x+2\)
Answer & Solution
Correct Answer
(B) \(y=x+2\)
Step-by-step Solution
Detailed explanation
Given, \(y^{2}=8 x\)
\(
\therefore 4 a=8 \Rightarrow a=2
\)
Any tangent of parabola is,
\(y =m x+\frac{a}{m}\)
\(\Rightarrow m x-y+\frac{2}{m}=0\)
If it is a tangent to the circle \(x^{2}+y^{2}=2\), then perpendicular from centre \((0,0)\) is equal to radius \(\sqrt{2}\).
\(\therefore \frac{\frac{2}{m}}{\sqrt{m^{2}+1}}=\sqrt{2} \)
\( \Rightarrow \frac{4}{m^{2}}=2\left(m^{2}+1\right) \)
\( \Rightarrow m^{4}+m^{2}-2=0 \)
\( \Rightarrow \left(m^{2}+2\right)\left(m^{2}-1\right)=0\)
\(\Rightarrow\)
\(m=\pm 1\)
Hence, the common tangents are \(y=\pm(x+2)\).
\(
\therefore 4 a=8 \Rightarrow a=2
\)
Any tangent of parabola is,
\(y =m x+\frac{a}{m}\)
\(\Rightarrow m x-y+\frac{2}{m}=0\)
If it is a tangent to the circle \(x^{2}+y^{2}=2\), then perpendicular from centre \((0,0)\) is equal to radius \(\sqrt{2}\).
\(\therefore \frac{\frac{2}{m}}{\sqrt{m^{2}+1}}=\sqrt{2} \)
\( \Rightarrow \frac{4}{m^{2}}=2\left(m^{2}+1\right) \)
\( \Rightarrow m^{4}+m^{2}-2=0 \)
\( \Rightarrow \left(m^{2}+2\right)\left(m^{2}-1\right)=0\)
\(\Rightarrow\)
\(m=\pm 1\)
Hence, the common tangents are \(y=\pm(x+2)\).
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