Q.Find the equations of the tangents to the hyperbola 3x2−4y2=12 which are
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Perpendicular Slopes Condition
Imagine two roads crossing at a right angle — that's perpendicular lines. The question is: how do their slopes relate?
The Intuition
Take a line with slope 2. That means for every 1 unit you move right, you go up 2 units — a fairly steep climb. Now picture a line perpendicular to it. If the first line is climbing steeply, the perpendicular line must be falling gently, or climbing very shallowly in the opposite direction.
Why? Because a right angle means the two lines "flip" the rise and run. One line's steepness becomes the other's shallowness, but in the opposite sign.
Try this: a line with slope 2 (rise 2, run 1). A perpendicular line should have rise 1 and run −2 — that gives slope −21. Notice: 2×(−21)=−1.
That's the pattern: the slopes are negative reciprocals of each other.
The Precise Statement
m1⋅m2=−1
Two non-vertical lines are perpendicular if and only if the product of their slopes is −1.
Equivalently: m2=−m11 (provided m1=0).
What About Vertical and Horizontal Lines?
A vertical line has undefined slope. A horizontal line has slope 0. Their product? Undefined — not −1. Yet they are clearly perpendicular.
The formula m1⋅m2=−1 only works when both slopes are defined (neither line is vertical). For a vertical line (x=c) and a horizontal line (y=d), they are perpendicular by definition — no slope calculation needed.
Quick Check
Are y=3x+2 and y=−31x−5 perpendicular?
3×(−31)=−1. Yes.
Are y=4x and y=4x+1 perpendicular?
4×4=16=−1. No — they're parallel.
Why It Works (A Short Proof)
›Proof
Two lines with slopes m1 and m2 make angles θ1 and θ2 with the positive x-axis, where tanθ1=m1 and tanθ2=m2. …
Write the hyperbola as 4x2−3y2=1; use y=mx±a2m2−b2 with m=1 (parallel) and m=−1 (perpendicular). …
With a2=4,b2=3: tangents parallel to y=x−7 are y=x±1; perpendicular ones are y=−x±1.
Write 3x2−4y2=12 as 4x2−3y2=1, so a2=4, b2=3.
A tangent of slope m to a2x2−b2y2=1 is y=mx±a2m2−b2.
(i) Parallel to y=x−7: slope m=1.
y=x±4(1)−3=x±1.
…
- CBSE 2026Set ANNUAL1 markQ.Find the slope of the line perpendicular to 3x - 4y + 10 = 0.
›Reveal solutionSolution
Find the slope of the given line, then take the negative reciprocal for the perpendicular slope.
Rewrite 3x−4y+10=0 in slope-intercept form:
4y=3x+10⟹y=43x+410
So the slope of the given line is m1=43.
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- CBSE 2025Set ANNUAL1 markMCQQ.When two lines L1 and L2 are perpendicular to each other then their slopes m1 and m2 are related as(a) m1×m2=1(b) m1×m2=−1(c) m2m1=∞(d) m1m2=∞
›Reveal solutionSolution
Two lines with slopes m1 and m2 are perpendicular if and only if m1m2=−1.
If a line L1 makes angle θ with the x-axis, its slope is m1=tanθ. A line perpendicular to it makes angle θ+90°, with slope: …
- CBSE 2023Set ANNUAL1 markMCQQ.If 3x−4y+7=0 and ax+6y+1=0 are perpendicular, then a=(a) 4(b) 5(c) 10(d) 8
›Reveal solutionSolution
Perpendicular lines have slopes whose product is −1; solving gives a=8.
Rewrite each line in slope form y=mx+c:
- 3x−4y+7=0⇒y=43x+47, slope m1=43
- ax+6y+1=0⇒y=−6ax−61, slope m2=−6a …
- CBSE 2023Set ANNUAL1 markMCQQ.Case study (continued — triangle with vertices A(3,1), M(2,-3), S(-3,3)): Slope of a line perpendicular to AS is(a) 31(b) −31(c) 3(d) -3
›Reveal solutionSolution
Find slope of AS, then take the negative reciprocal.
A(3,1), S(−3,3).
Slope of AS =−3−33−1=−62=−31.
…
- CBSE 2023Set ANNUAL1 markMCQQ.If m1 and m2 are the slopes of two perpendicular lines, then:(a) m2m1=−1(b) m1⋅m2=−1(c) m1=m2(d) m1+m2=0
›Reveal solutionSolution
The standard perpendicularity condition for two lines is m1m2=−1.
For two non-vertical lines with slopes m1 and m2, they are perpendicular if and only if m1⋅m2=−1 (one slope is the negative reciprocal of the other).
…
- CBSE 2022Set ANNUAL1 markMCQQ.If a line through the points (−2,6) and (4,8) is perpendicular to the line through the points (8,12) and (x,24), then(a) x=1(b) x=2(c) x=3(d) x=4
›Reveal solutionSolution
Find the first line's slope, use m1m2=−1 to get the second slope, then solve for x.
Slope of the line through (−2,6) and (4,8):
m1=4−(−2)8−6=62=31
For two lines to be perpendicular, m1m2=−1, so m2=−3.
Slope of the line through (8,12) and (x,24):
m2=x−824−12=x−812 …
- CBSE 2022Set ANNUAL1 markMCQQ.The co-ordinates of the orthocentre of the triangle whose sides are x=3, y=4 and 3x+4y=6 will be(a) (0,0)(b) (3,0)(c) (0,4)(d) (3,4)
›Reveal solutionSolution
The orthocentre is (3,4).
The sides x=3 (vertical) and y=4 (horizontal) are perpendicular and meet at (3,4), so the triangle has a right angle at that vertex.
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- CBSE 2018Set ANNUAL1 markQ.If lines y=mx+5 and 3x+5y=8 are mutually perpendicular, then find value of m.
›Reveal solutionSolution
Two lines are perpendicular when the product of their slopes is −1; find the slope of 3x+5y=8 and solve for m.
The line y=mx+5 has slope m1=m.
Rewrite 3x+5y=8 in slope form: 5y=−3x+8⇒y=−53x+58, so its slope is m2=−53.
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