KCET · Maths · Parabola
The equation of the tangent to the parabola \(y^{2}=4 x\) inclined at an angle of \(\frac{\pi}{4}\) to the positive direction of \(x\)-axis, is
- A \(x+y-4=0\)
- B \(x-y+4=0\)
- C \(x-y-1=0\)
- D \(x-y+1=0\)
Answer & Solution
Correct Answer
(D) \(x-y+1=0\)
Step-by-step Solution
Detailed explanation
Given, equation of parabola is \(y^{2}=4 x\).
Here, \(\quad a=1\)
Now, equation of tangent to the parabola in slope form is
\(\begin{aligned}
y &=m x+\frac{a}{m} \\
\Rightarrow \quad y &=m x+\frac{1}{m}...(i)
\end{aligned}\)
Also given that tangent to the parabola inclined at an angle of \(\frac{\pi}{4}\) to the \((+\mathrm{ve})\) direction of \(x\)-axis.
\(\begin{array}{lll}
\therefore & m=\tan \frac{\pi}{4}=1 & \\
\text { Then, } & y=(1) x+1 \\
\Rightarrow & x-y+1=0
\end{array} \quad \text { [from Eq. (i)] }\)
Here, \(\quad a=1\)
Now, equation of tangent to the parabola in slope form is
\(\begin{aligned}
y &=m x+\frac{a}{m} \\
\Rightarrow \quad y &=m x+\frac{1}{m}...(i)
\end{aligned}\)
Also given that tangent to the parabola inclined at an angle of \(\frac{\pi}{4}\) to the \((+\mathrm{ve})\) direction of \(x\)-axis.
\(\begin{array}{lll}
\therefore & m=\tan \frac{\pi}{4}=1 & \\
\text { Then, } & y=(1) x+1 \\
\Rightarrow & x-y+1=0
\end{array} \quad \text { [from Eq. (i)] }\)
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