Let P and Q be any points on the curves (x – 1)² + (y + 1)² = 1 and y = x², respectively. The distance between P and Q is minimum for some vale of the abscissa of P in the interval

Option 1 - <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>

Option 2 - <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>

Option 3 - <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>

Option 4 - <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>,</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math>

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Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

Correct Option - 3

Detailed Solution:

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f $(\frac{1}{4}) f (\frac{1}{3}) < 0 \Rightarrow t \in (\frac{1}{4}, \frac{1}{3})$

Let P(1 – sin q, -1 + cos q) $∴$ slope of normal = slope of CP $- \frac{1}{2 t} = \frac{c o s θ}{- s i n θ} \Rightarrow 2 t$ Þ = tan q according to question $x = 1 - s i n θ = 1 - \frac{2 t}{\sqrt[]{1 + 4 t^{2}}} = g (t) ∴ g (t) = 1 - \frac{2 t}{\sqrt[]{1 + 4 t^{2}}}$ ,

Þ g'(t) < 0 Þ g(t) is decreasing function in $t \in (\frac{1}{4}, \frac{1}{3}) \Rightarrow g (t) \in (0.440, 0.485) \in (\frac{1}{4}, \frac{1}{2})$

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Similar Questions for you

The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle $\frac{π}{4}$ at the origin, is equal to:

Alok Kumar Singh

Let A, A’ be (, 2) AB and A’B subtends $\frac{π}{4}$ angle at (0, 0) slope of OA = $\frac{2}{α}$

slope of OB = $\frac{3}{2}$

$t a n \frac{π}{4} = | \frac{\frac{2}{α} - \frac{3}{2}}{1 + \frac{2}{α} \cdot \frac{3}{2}} |$

$\Rightarrow 1 + \frac{3}{α} = \pm (\frac{4 - 3 α}{2 α})$

$\Rightarrow \frac{α + 3}{α} = \pm (\frac{4 - 3 α}{2 α}) \Rightarrow α = 10, - \frac{2}{5}$

now distance between A’A, (10, 2) & $(\frac{- 2}{5}, 2) i s \frac{52}{5}$

Vishal Baghel

Slope of AH = $\frac{a + 2}{1}$ slope of BC = $- \frac{1}{p} ∴ p = a + 2$ $∴ C (\frac{18 p - 30}{p + 1}, \frac{15 p - 33}{p + 1})$

slope of HC = $\frac{16 p - p^{2} - 31}{16 p - 32}$

slope of BC × slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

A point P moves so that the sum of squares of its distances from the points (1, 2) and (2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A , B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to:

Vishal Baghel

Let point P : (h, k)

Therefore according to question, ${(h - 1)}^{2} + {(k - 2)}^{2} + {(h + 2)}^{2} + {(k - 1)}^{2} = 14$

$∴$ locus of P (h, k) is $x^{2} + y^{2} + x - 3 y - 2 = 0$

Now intersection with x – axis are $x^{2} + x - 2 = 0 \Rightarrow x = - 2, 1$

Now intersection with y – axis are $y^{2} - 3 y - 2 = 0 \Rightarrow y = \frac{3 \pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2} (| x_{1} | + | x_{2} |) (| y_{1} | + | y_{2} |) = \frac{1}{2} \times 3 \times \sqrt[]{17} = \frac{3 \sqrt[]{17}}{2}$