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Point Form of Tangent: Ellipse

If ti is th...

Question

If $t_{i}$ is the length of tangent to the circle $x^{2} + y^{2} + 2 g_{i} x + 5 = 0, w h e r e i = 1, 2, 3$ from any point and $g_{1}, g_{2}, g_{3}$ are in A.P. and $A_{i} \equiv (g_{i}, t_{1}^{2})$ , then

A

A_{1}, A_{2}, A_{3}

are collinear

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B

t_{1}^{2}, t_{2}^{2}, t_{3}^{2}

are in GP

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C

A_{1}, A_{2}, A_{3}

form equilateral triangle

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D

A_{1}, A_{2}, A_{3}

form right angle triangle

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Solution

The correct option is A $A_{1}, A_{2}, A_{3}$ are collinear

Given, $t_{i}$ is the length of tangent to the circle $x^{2} + y^{2} + 2 g_{i} x + 5 = 0$ from any point $g_{1}, g_{2}, g_{3}$ are in AP
$A_{i} = (g_{i}, t_{i}^{2})$
$x^{2} + y^{2} + 2 g_{i} x + 5 = 0 (x + g_{i})^{2} + y^{2} + 5 - g_{i}^{2} = 0$
Centre $(- g_{1}, 0) r = \sqrt{g^{2} - 5}$
Centre lies on $X$ -axis
If $g_{i} + g_{i + 1} > c_{i} c_{i + 1}$
also,
$2 g_{2} = g_{1} + g_{3} g_{2} - g_{1} = g_{3} - g_{2} = d$
Let three point from which the tangent are drawn
$(g_{2}, 0)$ on $1^{s t}$ circle
$(g_{3}, 0)$ on $2^{n d}$ circle
$(g_{2}, 0)$ on $3^{r d}$ circle
$t_{1} = \sqrt{{(g_{2} + g_{1})}^{2} + 5 - {g_{1}}^{2}} = 2 g_{1} g_{2} + 5 + {g_{2}}^{2} t_{2} = \sqrt{{(g_{3} + g_{2})}^{2} + 5 - {g_{2}}^{2}} = 2 g_{2} g_{3} + 5 + {g_{3}}^{2} t_{3} = \sqrt{{(g_{1} + g_{3})}^{2} + 5 - {g_{3}}^{2}} = 2 g_{1} g_{3} + 5 + {g_{1}}^{2} A_{1} = (g_{1}, 2 g_{1} g_{2} + 5 + {g_{2}}^{2}) A_{2} = (g_{2}, 2 g_{2} g_{3} + 5 + {g_{3}}^{2}) A_{3} = (g_{3}, 2 g_{1} g_{3} + 5 + {g_{1}}^{2})$
If $A_{1}, A_{2}, A_{3}$ are collinear, Area $(A_{1} A_{2} A_{3}) = 0$
$[x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0 g_{1} [2 g_{3} (g_{2} - g_{1}) + (g_{3} - g_{1}) (g_{3} + g_{1})] + g_{2} [2 g_{1} (g_{3} - g_{2}) + (g_{1} - g_{2}) (g_{1} + g_{2})] + g_{3} [2 g_{2} (g_{1} - g_{3}) + (g_{2} - g_{3}) (g_{2} + g_{3})] = 0 g_{1} [2 d g_{3} + 2 d (2 g_{2})] + g_{2} [2 d g_{1} - d (2 g_{2} - d)] + g_{3} [2 g_{2} (- 2 d) + (- d) (g_{2} + d)] = 0$

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Similar questions

C_{i}

is the centre of the circle

x^{2} + y^{2} + 2 g_{i} x + 5 = 0

t_{i}

is the length of the tangent from any point to this circle,

i = 1, 2, 3

then the points

(g_{1}, t_{1}^{2})

(g_{2}, t_{2}^{2})

(g_{3}, t_{3}^{2})

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. and

a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. and

a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

a_{1}, a_{2}, a_{3}, . . . a_{n}

are in H.P then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

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