MHT CET · Maths · Parabola
The equation of the tangent to the parabola \(y^2=8 x\), which is parallel to the line \(4 x-y+3=0\) is
- A \(2 x-8 y+1=0\)
- B \(8 x-2 y+1=0\)
- C \(8 x+2 y+1=0\)
- D \(2 x-8 y-1=0\)
Answer & Solution
Correct Answer
(B) \(8 x-2 y+1=0\)
Step-by-step Solution
Detailed explanation
\(y^2=8 x\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=8 \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4}{y}
\end{aligned}\)
Slope of the line \(4 x-y+3=0\) is 4 .
Since the tangent is parallel to \(4 x-y+3=0\), their slopes are equal
\(\begin{aligned}
\therefore \quad & \frac{4}{y}=4 \\
& \Rightarrow y=1
\end{aligned}\)
When \(y=1, x=\frac{1}{8}\)
\(\therefore \quad\) Equation of the tangent at \(\left(\frac{1}{8}, 1\right)\) is
\(\begin{aligned}
& y-1=4\left(x-\frac{1}{8}\right) \\
& \Rightarrow 2 y-2=8 x-1 \\
& \Rightarrow 8 x-2 y+1=0
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=8 \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4}{y}
\end{aligned}\)
Slope of the line \(4 x-y+3=0\) is 4 .
Since the tangent is parallel to \(4 x-y+3=0\), their slopes are equal
\(\begin{aligned}
\therefore \quad & \frac{4}{y}=4 \\
& \Rightarrow y=1
\end{aligned}\)
When \(y=1, x=\frac{1}{8}\)
\(\therefore \quad\) Equation of the tangent at \(\left(\frac{1}{8}, 1\right)\) is
\(\begin{aligned}
& y-1=4\left(x-\frac{1}{8}\right) \\
& \Rightarrow 2 y-2=8 x-1 \\
& \Rightarrow 8 x-2 y+1=0
\end{aligned}\)
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