Question

The normal to the curve \(x^2 = 4y\) passing \((1, 2)\) is

• A. \(x+y=3\)
• B. \(x-y=3\)
• C. \(x+y=1\)
• D. \(x-y=1\)

Answer 1

The Correct Option is A

The correct answer is A:\(x+y=3\)
The equation of the given curve is \(x^2 = 4y\). Differentiating with respect to \(x\), we have:
\(2x=4.\frac{dy}{dx}\)
\(\frac{dy}{dx}=\frac{x}{2}\)
The slope of the normal to the given curve at point \((h, k)\) is given by,
\(\frac{-1}{\frac{dy}{dx}]_{(h,k)}}=\frac{-2}{h}\)
∴Equation of the normal at point \((h, k)\) is given as:
\(y-k=\frac{-2}{h}(x-h)\)
Now. it is given that the normal passes through the point \((1,2)\) Therefore, we have:
\(2-k=\frac{-2}{h}(1-h) or k=2+\frac{2}{h}(1-h) ....(i)\)
Since \((h, k)\) lies on the curve \(x^2 = 4y\), we have \(h^2 = 4k\)
\(k=\frac{h^2}{4}\)
From equation (i), we have:
\(\frac{h^2}{4}=2+\frac{2}{h}(1-h)\)
\(\frac{h^2}{4}=2h+2=2h=2\)
\(h^3=8\)
\(h=2\)
\(k=\frac{h^2}{4}=k=1\)
Hence, the equation of the normal is given as:
\(y=1=\frac{-2}{2}(x-2)\)
\(y-1=-(x-2)\)
\(x+y=3\)
The correct answer is A.