The point(s) at which the tangents to the cur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the tangent at point $P(x_1, y_1)$ intersect the $X$-axis at $A(a, 0)$ and the $Y$-axis at $B(0, -2a)$ (since the segment on $OX$ is half that

Vedclass · Accepted Answer

$(3, -15)$

Vedclass · Answer

(B) Let the tangent at point $P(x_1, y_1)$ intersect the $X$-axis at $A(a, 0)$ and the $Y$-axis at $B(0, -2a)$ (since the segment on $OX$ is half that on $OY$,and $B$ is on the negative $Y$-axis).The slope of the tangent is $m = \frac{0 - (-2a)}{a - 0} = \frac{2a}{a} = 2$.Given $y = x^3 - 3x^2 - 7x + 6$,the slope at any point is $\frac{dy}{dx} = 3x^2 - 6x - 7$.Equating the slope to $2$: $3x_1^2 - 6x_1 - 7 = 2 \implies 3x_1^2 - 6x_1 - 9 = 0 \implies x_1^2 - 2x_1 - 3 = 0$.Solving for $x_1$: $(x_1 - 3)(x_1 + 1) = 0$,so $x_1 = 3$ or $x_1 = -1$.If $x_1 = 3$,$y_1 = (3)^3 - 3(3)^2 - 7(3) + 6 = 27 - 27 - 21 + 6 = -15$. Point is $(3, -15)$.If $x_1 = -1$,$y_1 = (-1)^3 - 3(-1)^2 - 7(-1) + 6 = -1 - 3 + 7 + 6 = 9$. Point is $(-1, 9)$.Checking the intercept condition for $P(-1, 9)$: The tangent equation is $y - 9 = 2(x + 1) \implies y = 2x + 11$. $X$-intercept $A$ is $(-5.5, 0)$,$Y$-intercept $B$ is $(0, 11)$. The length on $OX$ is $5.5$ (negative side),which does not satisfy the condition.Checking $P(3, -15)$: The tangent equation is $y + 15 = 2(x - 3) \implies y = 2x - 21$. $X$-intercept $A$ is $(10.5, 0)$,$Y$-intercept $B$ is $(0, -21)$. The length on $OX$ is $10.5$,and on $OY$ is $21$. Since $10.5 = \frac{1}{2}(21)$,this point satisfies the condition.Thus,the correct point is $(3, -15)$.

The point$(s)$ at which the tangents to the curve $y = x^3 - 3x^2 - 7x + 6$ cut off a line segment on the positive semi-axis $OX$ that is half the length of the segment cut off on the negative semi-axis $OY$ are given by:

Similar Questions

If the tangent to the curve $y = f(x)$ at a point is perpendicular to the line $2x - 3y = 5$,then the value of $\frac{dy}{dx}$ at that point is:

The equation of a straight line passing through the point $(3, 6)$ and cutting the curve $y = \sqrt{x}$ orthogonally is

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $y$-axis.

The angle between the curves $x^2=8y$ and $xy=8$ is

The points on the curve $y^2 = \frac{x^3}{9}$,where the normal to the curve makes equal intercepts with the axes,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module