MHTCET 11 May 2024 Shift 1 Paper

Concept:For a parametric curve, eliminate the parameter to find the slope, then use the negative reciprocal to write the normal's equation.Explanation:Given $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$.Since $x=\sqrt{t}$, we get $t=x^2$ and $\sqrt{t}=x$.Substitute into $y$: $y=x^2-\frac{1}{x}$.At $t=4$, the point is $x=2$ and $y=4-\frac{1}{2}=\frac{7}{2}$.Differentiate: $\frac{dy}{dx}=2x+\frac{1}{x^2}$.At $x=2$, the tangent slope is $\frac{dy}{dx}=2(2)+\frac{1}{4}=\frac{17}{4}$.The slope of the normal is the negative reciprocal: $m=-\frac{4}{17}$.Using point-slope form through $\left(2,\frac{7}{2}\right)$: $y-\frac{7}{2}=-\frac{4}{17}(x-2)$.Multiply by $17$: $17y-\frac{119}{2}=-4x+8$.Multiply by $2$: $34y-119=-8x+16$.Rearranging gives $8x+34y=135$.Answer:The correct option is D: $8x+34y=135$.