Question

The equation of tangent of the curve \[ y=\sqrt{9-2x^2} \] at the point where the ordinate and abscissa are equal is

• A. \(2x+y-3\sqrt{3}=0\)
• B. \(2x+y+3\sqrt{3}=0\)
• C. \(2x-y-3\sqrt{3}=0\)
• D. \(2x-y+3\sqrt{3}=0\)

Hint:

When ordinate equals abscissa, simply use \[ y=x \] to find the required point on the curve.

Answer 1

The Correct Option is A

Step 1: Rewrite the curve in implicit form.
$ y=\sqrt{9-2x^2} \implies 2x^2+y^2=9 $ (an ellipse).
Step 2: Find the point where ordinate = abscissa.
$ y=x $: $ 2x^2+x^2=9 \implies x=\sqrt{3} $. Point: $ (\sqrt{3},\sqrt{3}) $.
Step 3: Differentiate the ellipse implicitly.
\[ 4x+2y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{2x}{y} \]
Step 4: Evaluate the slope at $ (\sqrt{3},\sqrt{3}) $.
\[ m = -\frac{2\sqrt{3}}{\sqrt{3}} = -2 \]
Step 5: Write the tangent equation.
\[ y-\sqrt{3} = -2(x-\sqrt{3}) \implies 2x+y = 3\sqrt{3} \]
Step 6: Verify and state the answer.
At $ x=\sqrt{3} $: $ 2\sqrt{3}+\sqrt{3}=3\sqrt{3} $. Correct. \[ \boxed{2x+y-3\sqrt{3}=0} \]