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Chapter 15: Problem 1388

If the line \(\mathrm{x}-1=0\) is the directrix of the parabola \(\mathrm{y}^{2}-\mathrm{kx}+8=0\) then one of the values of \(\mathrm{k}\) is (a) 4 (b) \((1 / 8)\) (c) \((1 / 4)\) (d) 8

Short Answer

Expert verified
One of the values of k is 4, which corresponds to option (a).

Step by step solution

01

1. Determine the general form of a parabola with a vertical axis of symmetry.

The general form of a parabola with a vertical axis of symmetry can be written as: \[y^2 = 4ax\]
02

2. Write the given parabola in vertex form.

The given equation of the parabola is: \[y^2 - kx + 8 = 0\] To write the given equation in vertex form, we need to make it look like the general form. So, we can rewrite the equation as: \[y^2 = kx - 8\]
03

3. Determine the vertex and focus of the parabola.

The vertex form of the given parabola is: \[y^2 = kx - 8\] Comparing the given equation with the general equation, we get 4a = k. The vertex of the parabola is at the origin (0,0), and the distance from the vertex to the focus is a. Therefore, the focus of the parabola is at (a, 0).
04

4. Use the directrix' equation to find the distance between the vertex and the focus.

The given equation of the directrix is x - 1 = 0, or x = 1. The distance from the vertex to the directrix is the same as the distance from the vertex to the focus. In this case, it is 1 unit since the vertex is located at the origin.
05

5. Use the distance between the vertex and the focus to find the value of k.

From the earlier steps, we know that the distance from the vertex to the focus (a) is 1 unit; therefore, 4a = k, and we can substitute a with the distance between the vertex and the focus: \[k = 4(1) = 4\] Hence, one of the values of k is 4. The correct answer is (a) 4.

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Most popular questions from this chapter

The area of the triangle formed by any tangent to the hyperbola \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)-\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) with its asymptotes is (a) \(a b\) (b) \(4 \mathrm{ab}\) (c) \(a^{2} b^{2}\) (d) \(4 \mathrm{a}^{2} \mathrm{~b}^{2}\)

Let \(E\) be the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1\) and \(C\) be the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\). Let \(\mathrm{P}\) and \(\mathrm{Q}\) be the point \((1,2)\) and \((2,1)\) respe. Then (a) P lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\) (b) P lies inside both \(\mathrm{C}\) and \(\mathrm{E}\) (c) \(Q\) lies outside both \(\mathrm{C}\) and \(\mathrm{E}\) (d) Q lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\)

If the tangents are drawn to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=12\) at the point where it meets the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-5 \mathrm{x}+3 \mathrm{y}-2=0\), then the point of intersection of these tangent is (a) \((6,-6)\) (b) \([6,(18 / 5)]\) (c) \([-6,(18 / 5)]\) (d) \([6,(18 / 5)]\)

If angle between asymptote of hyperbola \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)-\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) is \(45^{\circ} .\) The value of eccentricity is \(\ldots \ldots\) (a) \(\sqrt{(4 \pm 2 \sqrt{2})}\) (b) \(\sqrt{(4+2 \sqrt{2})}\) (c) \(\sqrt{(4-2 \sqrt{2})}\) (d) none of these

Let \(\mathrm{C}\) be the centre of the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-20=0\). If the tangents at the point \(\mathrm{A}(1,7)\) and \(\mathrm{B}(4,-2)\) on the circle meet at point \(\mathrm{D}\). Then area of the quadrilateral \(\mathrm{ABCD}\) is \(\ldots \ldots\) (a) 150 sq. units (b) 100 sq. units (c) 75 sq. units (d) 50 sq. units
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