If the tangent drawn at a point P on the curv | vedclass

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(B) Let the point $P(h, k)$ be on the curve $y=3x^2-5x+7$. The slope of the tangent at point $P$ is given by $\frac{dy}{dx} = 6x-5$. At point $P(h, k)

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let the point $P(h, k)$ be on the curve $y=3x^2-5x+7$. The slope of the tangent at point $P$ is given by $\frac{dy}{dx} = 6x-5$. At point $P(h, k)$, the slope is $6h-5$. The chord joins the points $(1, y_1)$ and $(2, y_2)$. For $x=1$, $y_1 = 3(1)^2 - 5(1) + 7 = 5$. For $x=2$, $y_2 = 3(2)^2 - 5(2) + 7 = 12 - 10 + 7 = 9$. The slope of the chord is $\frac{y_2-y_1}{2-1} = \frac{9-5}{1} = 4$. Since the tangent is parallel to the chord, their slopes are equal: $6h-5 = 4$. $6h = 9$. $h = \frac{9}{6} = \frac{3}{2}$. Thus, the $x$-coordinate of point $P$ is $\frac{3}{2}$.

If the tangent drawn at a point $P$ on the curve $y=3x^2-5x+7$ is parallel to its chord joining the points $(1, y_1)$ and $(2, y_2)$ on it, then the $x$-coordinate of the point $P$ is

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