Hyperbola Class 11 OP Malhotra Exe-25B ISC Maths Ch-25 Solutions. In this article you would learn about General Conic. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Hyperbola Class 11 OP Malhotra Exe-25B ISC Maths Solutions Ch-25
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-25 | Hyperbola |
| Writer | O.P. Malhotra |
| Exe-25(B) | General Conic |
General Conic.
Hyperbola Class 11 OP Malhotra Exe-25B ISC Maths Ch-25 Solutions.
Que-1: Find the tangent to the parabola y2 = 16x, making an angle of 45° with the x-axis.
Sol: The eqn. of tangent to parabola be
y = mx + am …(1)
Here m = tan 45° = 1
On comparing y² = 16x with y² = 4ax ;
we have, 16 = 4a
⇒ a = 4
Thus eqn. (1) reduces to ; y = x + 4
which is the required eqn. of tangent to given parabola.
Que-2: A tangent to the parabola y2 = 16x makes an angle of 60° with the x-axis. Find its point of contact.
Sol: We know that, the line y = mx + c may touch the parabola y2 =4ax then the point of contact be given by (a/m², 2a/m),
On comparing y² = 16x with y² = 4ax
we have, 16 = 4a ⇒ a = 4
and m = tan 60° = √3
∴ required point of contact be (4/3, (2×4)/√3) i.e. (4/√3, 8/√3)
Que-3: (i) Find the equations of the tangents to the parabola y2 = 6x which pass through the point (3/2, 5)
(ii) Find the equations of the tangents to the parabola y2 + 12x = 0 from the point (3, 8).
Sol: (i) On comparing y² = 6x with y² = 4ax; we have
4a = 6 ⇒ a = 3/2
We know that, eqn. of any tangent to parabola y² = 4ax be given by
y = mx + (a/m) …(1)
Thus required eqn. of tangent to parabola y² = 6x be given by
y = mx + (3/2m) …(2)
Now eqn. (2) passes through the point (32,5).
5 = (3/2) m + (3/2m)
⇒ 10m = 3m2 + 3
⇒ 3m2 – 10m + 3 = 0
⇒ (m – 3) (3m – 1) = 0
⇒ m = 3, 1/3
putting the values of m in eqn. (2); we have
y = 3x + {3/(2×3)}
⇒ y = 3x + (1/2)
⇒ 2y = 6x + 1 …(3)
and y = (1/3x) + {(3×3)/(2×1)}
⇒ y = (x/3) + (9/2)
⇒ 6y = 2x + 27 …(4)
Thus eqn. (3) and eqn. (4) are the required eqns. of tangents to given parabola.
(ii) Given eqn. of parabola be
y² = -12x …(1)
On comparing eqn. (1) with y2 = 4ax
we have, 4a = – 12 ⇒ a = – 3
We know that, eqn. of any tangent to parabola y² = 4ax be given by
y = mx + (a/m)
Thus eqn. of any tangent to given parabola (1) be given by
y = mx – (3/m)
Now eqn. (2) passes through the point (3, 8).
8 = 3m – (3/m) ⇒ 3m² – 8m – 3 = 0
⇒ m = {8±√(64+36)}/6 = (8±10)/6 = 3, 1/3
putting m = 3 in eqn. (2) ; we have
y = 3x – 1 …(3)
putting m = –1/3 in eqn. (2) ; we have
y = –1/3 + 9 ⇒ 3y = -x + 27 …(4)
Thus, eqn. (3) and eqn. (4) gives the required tangents to given parabola.
Que-4: Show that the line 12y – 20x – 9 = 0 touches the parabola y2 = 5x.
Sol: Given eqn. of parabola be
y² = 5x …(1)
On comparing eqn. (1) with y² = 4ax ; we have
4a = 5
⇒ a = 5/4
Given eqn. of line can be written as ;
y = (20x/12) + (9/12)
⇒ y = (5x/3) + (3/4) …(2)
Comparing eqn. (2) with y = mx + c
we have, m = 5/3 and c = 3/4
Here, am = {(5/4)/(5/3)} = 3/4 = c
Thus line (2) touches the parabola (1).
Que-5: Show that the line x + y = 1 touches the parabola y = x – x2.
Sol: Given eqn. of line be
x + y = 1 …(1)
and eqn. of parabola be y = x – x2 …(2)
From (1); y = 1 – x, putting in eqn. (2); we have
1 – x = x – x2
⇒ x2 – 2x + 1 = 0
⇒ (x – 1)2 = 0 …(3)
i.e. eqn. (3) have equal roots.
Thus eqn. (1) touches eqn. (2).
Que-6: Show that the line x + ny + an2 = 0 touches the parabola y2 = 4ax and find the point of contact.
Sol: Given eqn. of line be
x + ny + an2 = 0 …(1)
y2 = 4ax …(2)
From (1); y = (−x−an²)/n
putting the value of y in eqn. (2) ; we have
[(−x−an²)/n]² = 4ax
which is quadratic in x and have equal roots. Thus line (1) touches parabola (2).
∴ from (3); x = an2
from (1); ny + 2an2 = 0 ⇒ y = – 2an
Hence the required point of contact be (an2, – 2an).
Que-7: Find the tangents to the ellipse x2 + 9y2 = 3, which are (i) parallel (ii) perpendicular to the line 3x + 4y = 9.
Sol: (i) Given eqn. of ellipse be
x² + 9y² = 3 ⇒ (x²/3) + {y²/(1/3)} = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with (x²/a²) + (y²/b²) = 1
we have a² = 3 and b² = 1/3
eqn. of given line be
3x + 4y – 9 = 0 …(2)
∴ slope of line (2) = –3/4
∴ slope of line || to line (2) = –3/4 = m
The eqns. of tangents to ellipse (1) be
(ii) slope of line ⊥ to line (2)
= {−1/(−3/4)} = 4/3 = m
∴ required eqns. of tangents to ellipse (1) be given by
Que-8: Find the equations of the tangents to the ellipse (x²/2) + (y²/7) = 1 that make an angle of 45° with the x-axis.
Sol: Given eqn. of ellipse be
(x²/2) + (y²/7) = 1 …(1)
which is a vertical ellipse.
On comparing eqn. (1) with
(x²/b²) + (y²/a²) = 1, a > b > o
We have a² = 7 ; b² = 2
Here m = tan 45° = 1
The eqns. of tangents to given ellipse (1) be given by
y = mx ± √{b²m²+a²}
⇒ y = x ± √(2+7)
⇒ y = x ± 3
Que-9: Find the equation of the tangents to the ellipse (x²/16) + (y²/9) = 1, which make equal intercepts on the axes.
Sol: Given eqn. of ellipse be,
(x²/16) + (y²/9) = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with (x²/a²) + (y²/b²) = 1
where a > b > 0
we have a² = 16 and b² = 9
Since tangents makes equal intercepts on axes
∴ slope of tangents = m = ± 1
Thus required eqns. of tangents to eqn. (1) be given by
y = mx ± √{a²m²+b²}
y = ± x ± √{16(1)²+9}
y = ± x ± 5
Que-10: Find the value of ‘c’ so that 2x -y + c = 0 may touch the ellipse x2 + 2y2 = 2.
Sol: Given eqn. of line be 2x -y + c = 0
⇒ y = 2x + c …(1)
and eqn. of given ellipse be x² + 2y² = 2
⇒ (x²/2) + (y²/1) = 1 …(2)
We know that the line y = mx + c touches the ellipse (x²/a²) + (y²/b²) = 1
if c = ± √(a²m²+b²)
if c = ± √(2×2²+1)
⇒ c = ± 3
[Here m = 2 ; a² = 2 ; b² = 1]
Que-11: Show that the line lx + my = 1 will touch the ellipse (x²/a²) + (y²/b²) = 1 if a²l² + b²m² = 1.
Sol: eqn. of given line be
lx + my = 1 …(1)
and eqn. of ellipse be (x²/a²) + (y²/b²) = 1 … (2)
From (1) ; y = (1−lx)/m
putting the value of y in eqn. (2); we have
(x²/a²) + (1/b²) [(1−lx)/m]² = 1
⇒ m2b2x2 + a2 (1 – lx)2 = a2b2m2
⇒ x2 [m2b2 + a2l2] – 2a2lx + a2 – a2b2m2 = 0 …(3)
Now eqn. (1) touches eqn. (2)
if roots of quadratic eqn. (3) are equal
if Discriminant = 0
if (- 2a2l)2 – 4 (b2m2 + a2l2) (a2 – a2b2m2) = 0
if 4a4l2 – 4a2b2m2 + 4a2b4m4 – 4a4l2 + 4a4b2l2m2 = 0
if 4a2b2m2 (b2m2 + a2l2 – 1) = 0
if b2m2 + a2l2 = 1 which is the required condition.
Que-12: Show that the following lines are tangents to the given hyperbola and determine the points of contact.
(i) x + 1 = 0, 4x2 – 3y2 = 4
(ii) x – 2y + 1 = 0, x2 – 6y2 = 3
Sol: Given eqn. of line be x + 1 = 0 ⇒ x = – 1 …(1)
and eqn. of given hyperbola be 4x² – 3y² = 4 i.e. (x²/1) – {y²/(4/3)} = 1
putting eqn. (1) in eqn. (2) ; we have
4 – 3x2 = 4 ⇒ y2 = 0 which is quadratic in y and gives equal roots and each root be 0.
Thus line (1) touches hyperbola (2).
Putting y = 0 in eqn. (1); x = – 1
Hence the required point of contact be (- 1, 0)
(ii) Given eqn. of line be x- 2y + 1 = 0 …(1)
and eqn. of hyperbola be x2 – 6y2 = 3 …(2)
From (1); y = x+12, putting the value of y in eqn. (2); we have
x² – 6[(x+1)/2]² = 3
⇒ x² – (3/2) (x + 1)² = 3
⇒ 2x2 – 3 (x2 + 2x + 1) = 6
⇒ -x2 – 6x – 9 = 0
⇒ (x – 3)2 = 0
Clearly eqn. (3) have equal roots.
Thus line (1) touches given hyperbola (2).
∴ from (3); x + 3 = 0 ⇒ x = – 3
∴ from (1); y = – 1
Thus, the required point of contact be (- 3, – 1).
Que-13: Find the equations of the tangents to the hyperbola 2x2 – 3y2 = 6, which re parallel to the Iine x + y – 2 = 0.
Sol: eqn. of given hyperbola be, 2x² – 3y² = 6 ⇒ (x²/3) – (y²/2) = 1
On comparing eqn. (1) with (x²/a²) – (y²/b²) = 1
we have a² = 3 and b² = 2
and eqn. of given line be x + y – 2 = 0
i.e. y = – x + 2 …(2)
On comparing eqn. (2) with y = mx + c
we have m = – 1 and c = 2
Thus the required eqns. of tangents to eqn. (1) be given by
y = mx ± √{a²m²−b²}
⇒ y =- x ± √{3(−1)²−2}
⇒ y = – x ± 1
⇒ x + y ± 1 = 0
Que-14: The tangents from P to the hyperbola (x²/a²) – (y²/b²) = 1 are mutually perpendicular, show that the locus of P is the circle x2 + y2 = a2 – b2.
Sol: Given eqn. of hyperbola be
(x²/a²) – (y²/b²) = 1 …(1)
Thus, eqn. of tangents to hyperbola (1) at P (x, y) be given by
y = mx ± √{a²m²−b²}
⇒ y – mx = ± √{a²m²−b²}
On squaring both sides ; we have
(y – mx)2 = a2m2 – b2
⇒ m2x2 +y2 – 2myx + b2 – a2m2 = 0
⇒ m2x2 – 2myx + (b2 – a2m2 + y2) = 0
⇒ m2(x2 – a2) – 2xym + b2 + y2 = 0 …(2)
eqn. (2) is a quadratic in m and hence two roots say m1 and m2
∴ product of roots = m1 m2 = – 1
{(y²+b²)/(x²-a²)} = -1
y2 + b2 = -x2 + a2
⇒ x2 +y2 = a2 – b2 which is the required locus of point P.
Que-15: Show that the straight line x + y = 1 touches the hyperbola 2x2 – 3y2 = 6. Also find the coordinates of the point of contact.
Sol: Given eqn. of line be
x + y =1 …(1)
and eqn. of given hyperbola be
2x2 – 3y2 = 6 …(2)
∴ from (1); y = 1 – x, putting in eqn. (3); we get
2x2 – 3 (1 – x)2 = 6
⇒ 2x2 – 3 (x2 + 1 – 2x) = 6
⇒ -x2 + 6x – 9 = 0
⇒ x2 – 6x + 9 = 0
⇒ (x – 3)2 = 0 …(3)
Thus eqn. (3) is a quadratic in x and have two equal roots i.e. x = + 3, + 3.
∴ eqn. (1) touches hyperbola (2).
putting x = + 3 in eqn. (1); we have y = -2.
∴ required point of contact be (+ 3, – 2).
Que-16: Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 144, which are perpendicular to the line 6x + 5y = 21.
Sol: Given eqn. of hyperbola be
4x² – 9y² = 144 ⇒ (x²/36) – (y²/16) = 1 …(1)
On comparing eqn. (1) with x2a2 – y2b2 = 1
we have, a² = 36 and b² = 16
slope of given line 6x + 5y – 21 = 0 be −6/5
∴ slope of line ⊥ to given line = {−1/(−6/5)} = 5/6 = m
Thus, eqns. of tangents to hyperbola (1) be given by
y = mx ± √{a²m²−b²}
⇒ y = 5/6 x ± √{36×(25/36)−16}
⇒ y = 5/6 x ± 3
⇒ 6y = 5x ± 18
⇒ 5x – 6y ± 18 = 0
–: End of Hyperbola Class 11 OP Malhotra Exe-25B ISC Maths Ch-25 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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