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For which of the hyperbolas, can we have...

For which of the hyperbolas, can we have more than one pair of perpendicular tangents? `(x^2)/4-(y^2)/9=1` (b) `(x^2)/4-(y^2)/9=-1` `x^2-y^2=4` (d) `x y=44`

A

`(x^(2))/(4)-(y^(2))/(9)=1`

B

`(x^(2))/(4)-(y^(2))/(9)=-1`

C

`x^(2)-y^(2)=4`

D

`xy=44`

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The correct Answer is:
To determine which hyperbolas can have more than one pair of perpendicular tangents, we need to analyze the given equations using the properties of hyperbolas and their director circles. ### Step-by-Step Solution: 1. **Identify the General Form of Hyperbola**: The general form of a hyperbola is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] or \[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \] 2. **Understand the Condition for Perpendicular Tangents**: For a hyperbola, the intersection point of two perpendicular tangents lies on the director circle. The equation of the director circle is given by: \[ x^2 + y^2 = a^2 - b^2 \] This condition holds true when \( a^2 > b^2 \). 3. **Analyze Each Option**: - **Option (a)**: \(\frac{x^2}{4} - \frac{y^2}{9} = 1\) - Here, \( a^2 = 4 \) and \( b^2 = 9 \). Since \( 4 < 9 \), this hyperbola does not satisfy the condition for perpendicular tangents. - **Option (b)**: \(\frac{x^2}{4} - \frac{y^2}{9} = -1\) - This represents a hyperbola in the form \(\frac{y^2}{9} - \frac{x^2}{4} = 1\). Here, \( b^2 = 9 \) and \( a^2 = 4 \). Since \( 9 > 4 \), this hyperbola can have perpendicular tangents. - **Option (c)**: \(x^2 - y^2 = 4\) - This can be rewritten as \(\frac{x^2}{4} - \frac{y^2}{4} = 1\). Here, \( a^2 = 4 \) and \( b^2 = 4 \). Since \( a^2 = b^2 \), this hyperbola does not satisfy the condition for perpendicular tangents. - **Option (d)**: \(xy = 44\) - This is a rectangular hyperbola, which can be expressed in the form \(\frac{x^2}{44} - \frac{y^2}{44} = 1\). Here, \( a^2 = 44 \) and \( b^2 = 44 \). Since \( a^2 = b^2 \), this hyperbola does not satisfy the condition for perpendicular tangents. 4. **Conclusion**: The only hyperbola that can have more than one pair of perpendicular tangents is option (b): \[ \frac{x^2}{4} - \frac{y^2}{9} = -1 \]

To determine which hyperbolas can have more than one pair of perpendicular tangents, we need to analyze the given equations using the properties of hyperbolas and their director circles. ### Step-by-Step Solution: 1. **Identify the General Form of Hyperbola**: The general form of a hyperbola is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ...
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For which of the hyperbolas, can we have more than one pair of perpendicular tangents? (a) (x^2)/4-(y^2)/9=1 (b) (x^2)/4-(y^2)/9=-1 (c)x^2-y^2=4 (d) x y=44

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (x^2+y^2)^2=4x y (b) (x^2-y^2)=1/9 (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (a) (x^2+y^2)^2=4x y (b) (x^2-y^2)=1/9 (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

If x=9 is the chord of contact of the hyperbola x^2-y^2=9 then the equation of the corresponding pair of tangents is (A) 9x^2-8y^2+18x-9=0 (B) 9x^2-8y^2-18x+9=0 (C) 9x^2-8y^2-18x-9=0 (D) 9x^2-8y^2+18x+9=0

If x=9 is the chord of contact of the hyperbola x^2-y^2=9 then the equation of the corresponding pair of tangents is (A) 9x^2-8y^2+18x-9=0 (B) 9x^2-8y^2-18x+9=0 (C) 9x^2-8y^2-18x-9=0 (D) 9x^2-8y^2+18x+9=0

If x=9 is the chord of contact of the hyperbola x^2-y^2=9 then the equation of the corresponding pair of tangents is (A) 9x^2-8y^2+18x-9=0 (B) 9x^2-8y^2-18x+9=0 (C) 9x^2-8y^2-18x-9=0 (D) 9x^2-8y^2+18x+9=0

The locus of the foot of the perpendicular from the center of the hyperbola x y=1 on a variable tangent is (a) (x^(2)+y^(2))^(2)=4xy (b) (x^2-y^2)=1/9 (c) (x^2-y^2)=7/(144) (d) (x^2-y^2)=1/(16)

If the angle between the asymptotes of hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 is 120^0 and the product of perpendiculars drawn from the foci upon its any tangent is 9, then the locus of the point of intersection of perpendicular tangents of the hyperbola can be (a) x^2+y^2=6 (b) x^2+y^2=9 (c) x^2+y^2=3 (d) x^2+y^2=18

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