MHT CET · Maths · Parabola
The equation to the line touching both the parabolas \(y^{2}=4 x\) and \(x^{2}=-32 y\), is
- A \(x+2 y+4=0\)
- B \(2 x+y-4=0\)
- C \(x-2 y-4=0\)
- D \(x-2 y+4=0\)
Answer & Solution
Correct Answer
(D) \(x-2 y+4=0\)
Step-by-step Solution
Detailed explanation
The equation of any tangent to the parabola \(y^{2}=4 x\) is
\(
y=m x+\frac{1}{m}
\)
This touches the parabola \(x^{2}=-32 y\), therefore the equation \(x^{2}=-32\left(m x+\frac{1}{m}\right)\) has equal roots
\(\therefore\) \(
(32 \mathrm{~m})^{2}=4\left(\frac{32}{\mathrm{~m}}\right)
\)
\(
\Rightarrow 8 m^{3}=1 \quad \Rightarrow \quad m=\frac{1}{2}
\)
On putting the value of \(m\) in Eq. (i), we get
\(
x-2 y+4=0
\)
\(
y=m x+\frac{1}{m}
\)
This touches the parabola \(x^{2}=-32 y\), therefore the equation \(x^{2}=-32\left(m x+\frac{1}{m}\right)\) has equal roots
\(\therefore\) \(
(32 \mathrm{~m})^{2}=4\left(\frac{32}{\mathrm{~m}}\right)
\)
\(
\Rightarrow 8 m^{3}=1 \quad \Rightarrow \quad m=\frac{1}{2}
\)
On putting the value of \(m\) in Eq. (i), we get
\(
x-2 y+4=0
\)
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