Question

Let 'O' be origin and P be a variable point on the conic $y = \sqrt{5x^2 + 7x}$ and if Q is the foot of perpendicular from P on x-axis then the value of $Lt_{OQ \to 0} \frac{(area \text{ } of \text{ } (\triangle OPQ))}{\Delta}$ (where $\Delta$ is the area bounded by the conic and x-axis and ordinate through P) is equal to k, then the value of 4k-2=

Answer 1

1

Answer 2

To find the value of $4k-2$, we first need to determine the value of $k$.

Given:

1. O is the origin (0,0).
2. P is a variable point $(x,y)$ on the conic $y = \sqrt{5x^2 + 7x}$.
3. Q is the foot of the perpendicular from P on the x-axis, so Q has coordinates $(x,0)$.
4. $k = Lt_{OQ \to 0} \frac{(area \text{ } of \text{ } (\triangle OPQ))}{\Delta}$.
5. $\Delta$ is the area bounded by the conic, the x-axis, and the ordinate through P.

Step 1: Determine the domain of the conic. For $y = \sqrt{5x^2 + 7x}$ to be real, $5x^2 + 7x \ge 0$. Factoring, $x(5x+7) \ge 0$. This inequality holds when $x \ge 0$ or $x \le -7/5$. The limit is $OQ \to 0$. Since $OQ = |x|$, this means $|x| \to 0$, which implies $x \to 0$. For $x \to 0$, we must consider $x > 0$ because if $x < 0$, then $x$ must be less than or equal to $-7/5$, which does not approach 0. So, we consider $x \to 0^+$. In this case, $x > 0$ and $y > 0$.

Step 2: Calculate the area of $\triangle OPQ$. The base of $\triangle OPQ$ is OQ, which lies on the x-axis. Its length is $OQ = x$. The height of $\triangle OPQ$ is PQ, which is perpendicular to the x-axis. Its length is $PQ = y$. Area of $\triangle OPQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} x y$. Substitute $y = \sqrt{5x^2 + 7x}$: Area of $\triangle OPQ = \frac{1}{2} x \sqrt{5x^2 + 7x}$.

Step 3: Calculate $\Delta$. $\Delta$ is the area bounded by the conic $y = \sqrt{5x^2 + 7x}$, the x-axis, and the ordinate through P (which is $x$). Since the conic passes through the origin ($y=0$ when $x=0$), the area is given by the definite integral from $0$ to $x$: $\Delta = \int_{0}^{x} \sqrt{5t^2 + 7t} \text{ } dt$.

Step 4: Evaluate the limit $k$. The expression for $k$ is: $k = Lt_{x \to 0^+} \frac{\frac{1}{2} x \sqrt{5x^2 + 7x}}{\int_{0}^{x} \sqrt{5t^2 + 7t} \text{ } dt}$. As $x \to 0^+$, the numerator approaches $\frac{1}{2} (0) \sqrt{0} = 0$. The denominator approaches $\int_{0}^{0} \sqrt{5t^2 + 7t} \text{ } dt = 0$. Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule. Let $f(x) = \frac{1}{2} x \sqrt{5x^2 + 7x}$ and $g(x) = \int_{0}^{x} \sqrt{5t^2 + 7t} \text{ } dt$. We need to find $f'(x)$ and $g'(x)$. By the Fundamental Theorem of Calculus, $g'(x) = \sqrt{5x^2 + 7x}$.

Now, calculate $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx} \left( \frac{1}{2} x \sqrt{5x^2 + 7x} \right)$ $f'(x) = \frac{1}{2} \left[ (1) \cdot \sqrt{5x^2 + 7x} + x \cdot \frac{d}{dx}(\sqrt{5x^2 + 7x}) \right]$ $f'(x) = \frac{1}{2} \left[ \sqrt{5x^2 + 7x} + x \cdot \frac{1}{2\sqrt{5x^2 + 7x}} \cdot (10x + 7) \right]$ To simplify, find a common denominator: $f'(x) = \frac{1}{2} \left[ \frac{2(5x^2 + 7x) + x(10x + 7)}{2\sqrt{5x^2 + 7x}} \right]$ $f'(x) = \frac{1}{4} \frac{10x^2 + 14x + 10x^2 + 7x}{\sqrt{5x^2 + 7x}}$ $f'(x) = \frac{1}{4} \frac{20x^2 + 21x}{\sqrt{5x^2 + 7x}}$.

Now, apply L'Hopital's Rule: $k = Lt_{x \to 0^+} \frac{f'(x)}{g'(x)} = Lt_{x \to 0^+} \frac{\frac{1}{4} \frac{20x^2 + 21x}{\sqrt{5x^2 + 7x}}}{\sqrt{5x^2 + 7x}}$ $k = Lt_{x \to 0^+} \frac{1}{4} \frac{20x^2 + 21x}{5x^2 + 7x}$ Factor out $x$ from the numerator and denominator: $k = Lt_{x \to 0^+} \frac{1}{4} \frac{x(20x + 21)}{x(5x + 7)}$ Since $x \to 0^+$, $x \ne 0$, so we can cancel $x$: $k = Lt_{x \to 0^+} \frac{1}{4} \frac{20x + 21}{5x + 7}$ Now, substitute $x=0$: $k = \frac{1}{4} \frac{20(0) + 21}{5(0) + 7} = \frac{1}{4} \frac{21}{7} = \frac{1}{4} \cdot 3 = \frac{3}{4}$.

Step 5: Calculate the value of $4k-2$. $4k - 2 = 4 \left( \frac{3}{4} \right) - 2 = 3 - 2 = 1$.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Identify coordinates: O(0,0), P(x,y), Q(x,0).
2. Area of $\triangle OPQ = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} x y$.
3. Area $\Delta = \int_0^x y \text{ } dt$.
4. Substitute $y = \sqrt{5x^2+7x}$ into both expressions.
5. The limit $OQ \to 0$ means $x \to 0^+$.
6. The limit $k = Lt_{x \to 0^+} \frac{\frac{1}{2} x \sqrt{5x^2+7x}}{\int_0^x \sqrt{5t^2+7t} dt}$ is of $\frac{0}{0}$ form.
7. Apply L'Hopital's Rule: $k = Lt_{x \to 0^+} \frac{\frac{d}{dx}(\frac{1}{2} x \sqrt{5x^2+7x})}{\frac{d}{dx}(\int_0^x \sqrt{5t^2+7t} dt)}$.
8. Calculate derivatives: Numerator derivative: $\frac{1}{4} \frac{20x^2+21x}{\sqrt{5x^2+7x}}$. Denominator derivative (by FTC): $\sqrt{5x^2+7x}$.
9. Substitute derivatives into the limit expression: $k = Lt_{x \to 0^+} \frac{\frac{1}{4} \frac{20x^2+21x}{\sqrt{5x^2+7x}}}{\sqrt{5x^2+7x}} = Lt_{x \to 0^+} \frac{1}{4} \frac{20x^2+21x}{5x^2+7x}$.
10. Simplify the expression by factoring out $x$: $k = Lt_{x \to 0^+} \frac{1}{4} \frac{x(20x+21)}{x(5x+7)} = Lt_{x \to 0^+} \frac{1}{4} \frac{20x+21}{5x+7}$.
11. Evaluate the limit by substituting $x=0$: $k = \frac{1}{4} \frac{21}{7} = \frac{3}{4}$.
12. Calculate $4k-2 = 4(\frac{3}{4}) - 2 = 3 - 2 = 1$.