The number of points on the curve y=2t^2+3t-5 | vedclass

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(D) The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t+3}{3t^2-8t-3}$.Since the normal is parallel to t

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$2$

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(D) The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t+3}{3t^2-8t-3}$.Since the normal is parallel to the $X$-axis,the tangent at that point must be perpendicular to the $X$-axis,which means the tangent is parallel to the $Y$-axis.Thus,the slope of the tangent $\frac{dy}{dx}$ must be undefined,or $\frac{dx}{dt} = 0$.Setting the denominator to zero: $3t^2-8t-3 = 0$.Solving the quadratic equation: $(3t+1)(t-3) = 0$.This gives $t = 3$ and $t = -\frac{1}{3}$.For these values of $t$,the slope of the tangent is undefined,implying the normal is parallel to the $X$-axis.Therefore,there are $2$ such points.

The number of points on the curve $y=2t^2+3t-5$ and $x=t^3-4t^2-3t$ such that the normals drawn at them on the curve are parallel to the $X$-axis is

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